TLO 1 Notes.pdf

MODULE 1 This module covers the fundamental concepts of forces. The force is an important factor in mechanics, and its effects on particles and rigid bodies will be discussed in this modul...

MODULE 1 This module covers the fundamental concepts of forces. The force is an important factor in mechanics, and its effects on particles and rigid bodies will be discussed in this module. In particular, you will learn how to resolve forces into components, as well as to reduce a system of forces into a simpler equivalent system. You will also learn how to apply vector algebra to solve problems involving forces. This module is divided into 2 units, and each unit is divided into 5 parts, namely, engage, explore, explain, elaborate, evaluate. UNIT 1. INTRODUCTORY CONCEPTS At the end of this unit you should be able to demonstrate knowledge on the basic concepts of force. ENGAGE Before studying this unit, take the test to determine how much you already know about forces. Direction: Choose the letter of the correct answer. Write your answer on the space provided before each number. 1. A force is a physical quantity having A. Both magnitude and direction B. Magnitude only C. Direction only D. None of the above 2. On Earth, which one of the following statements about weight is TRUE? A. An object’s weight always has a magnitude of m·g B. Weight always points perpendicular to the surface of contact. C. Weight is a scalar quantity D. An object’s weight depends on the velocity of the object. 3. Which one of the following is equal to a newton? A. kg/s2 B. kg·m/s2 C. kg·m/s D. kg2·m2/s2 4. Which of the following statements expresses Newton’s First Law? A. When an object experiences a net force, the net force (i.e., the vector sum of all the forces acting on it) is equal to the object’s mass times its acceleration. Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 5  B. The net work done on an object is equal to the change in its kinetic energy. C. If object 1 exerts a force on object 2, then object 2 exerts a force on object 1, and these two forces are equal in magnitude and opposite in direction. D. An object at rest remains at rest, and an object in motion continues in motion with constant velocity, unless acted upon by an external force. 5. A car pulls on a rope tied to a tree with a force of 1000 N. Which of the following is not true? A. The tree pulls on the rope with a force of 1000 N B. The tension in the rope is 1000 N. C. The tension in the rope is 2000 N. D. The rope pulls on the tree with a force of 1000 N. 6. The pulling force exerted by a stretched rope or cord on an object to which it’s attached is called A. Friction force B. Gravitational force C. Normal force D. Tension force 7. Refer to Fig. E1.7. A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20o and the man pulls upward with a force F whose direction makes an angle of 30o with the ramp. How large a force F is necessary for the component Fx parallel to the ramp to be 60.0 N? A. 30.00 N ↗ F B. 38.57 N ↗ C. 45.96 N ↗ D. 69.28 N ↗ Fig. E1.7 8. Referring to Problem 7, how large will the component Fy perpendicular to the ramp then be? A. 30.00 N ↖ B. 34.64 N ↖ +y C. 45.96 N↖ F = 1.50 kN D. 69.28 N ↖ 45o 9. Refer to Fig. E1.9. The component Fx of the 1.5 kN force is +x A. + 0.92 kN B. – 0.92 kN C. + 1.06 kN 60o D. – 1.06 kN +z Fig. E1.9 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 6  10. Refer to Fig. E9. The component Fz of the 1.5 kN force is A. – 0. 53 kN B. + 0.53 kN C. – 0. 75 kN D. + 0.75 kN Refer to the Answer Key and check your answers. What is your score? ______/10 Answer Key: B 10. A 9. B 8. D 7. D 6. C 5. D 4. B 3. A 2. A 1. EXPLORE Read and understand the prepared lessons and find out how much you have learned by answering the activities under the Elaborate and Evaluation sections. INTRODUCTION Mechanics is a branch of physical science which describes and predicts the conditions of rest or motion of bodies under the action of forces. Thus, engineering mechanics is the branch of engineering that applies the principles of mechanics to mechanical design (i.e., any design that must take into account the effect of forces). Engineering Mechanics Statics Dynamics - deals with the study of - deals with study of bodies bodies at rest. in motion. Kinematics Kinetics Figure 1.1. Classification of Engineering Mechanics Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 7  1.1 FUNDAMENTAL CONCEPTS AND AXIOMS Particle - A body with mass but with dimensions that can be neglected. Rigid body - Defined as a definite amount of matter the parts of which are fixed in position relative to one another. Mass - Invariant property of a body which measures its resistance to a change of motion. Force - The action exerted by one body upon another. - It is represented by a vector. P 30o Figure 1.2. Graphical representation of a force. External effect of a force - Manifested by a change in, or a tendency to change, the state of motion of the body upon which it acts. Internal effect of a force - Produce stress and deformation in the body. Characteristics of a Force 1. Magnitude 2. Position of its line of action 3. Direction (or sense) in which the force acts along its line of action. A change in any of these characteristics will result in a corresponding change in the external and internal effects of the force. Unit of force: SI system: newton (N) U.S. customary units: pound (lb) or kip (k) = 1000 lb Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 8  Principle of Transmissibility - States that the external effect of a force on a rigid body is the same for all points of application along its line of action. - The principle of transmissibility applies only to the external effect of a force on the same rigid body. = Figure 1.3. The effect of an external force on a rigid body remains unchanged if that force is moved along its line of action. 1.2 AXIOMS OF MECHANICS 1. The parallelogram law: The resultant of two forces is the diagonal formed on the vectors of these force. A R B Figure 1.4. Parallelogram law 2. Two forces are in equilibrium only when equal in magnitude, opposite in direction, and collinear in action. 3. A set of forces in equilibrium may be added to any system of forces without changing the effect of the original system. 4. Action and reaction forces are equal but oppositely directed. Force of A on B Force of B on A A B F F Figure 1.5. Action – reaction Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 9  Guide to solving problems: The solutions must be based on the axioms of mechanics. 1. Problem Statement: Includes given data, specification of what is to be determined, and a figure showing all quantities involved. 2. Free-Body Diagrams: Create separate diagrams for each of the bodies involved with a clear indication of all forces acting on each body. 3. Fundamental Principles: Apply the relevant principles in equation form. Make sure that the equations used in the computations are dimensionally homogeneous. The rules of algebra are applied to solve the equations for the unknown quantities. 4. Solution Check: Evaluate your answer by verifying that the units of the computed results are correct. Apply experience and physical intuition to assess whether results seem “reasonable”. 1.3 SYSTEM OF FORCES When several forces of various magnitude and direction act upon a body they are said to form system of forces. 1.Coplanar forces - the forces whose lines of action lie on the same plane. 2.Collinear forces - the forces whose lines of action lie on the same line 3.Concurrent forces - the forces whose lines of action intersect at one point. 4.Coplanar concurrent forces - the forces which meet at one point and their lines of action also lie on the same plane. 5. Coplanar non-concurrent forces - the forces which do not meet at one point, but their lines of action lie on the same plane. 6. Non-coplanar concurrent forces - the forces which meet at one point, but their lines of action do not lie on the same plane. 7. Non-coplanar non-concurrent forces - the forces which do not meet at one point and their lines of action do not lie on the same plane Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 10  Coplanar Concurrent (in plane) Parallel 2-dimensional Nonconcurrent System of General forces Noncoplanar Concurrent (in space) Parallel 3-dimensional Nonconcurrent General Figure 1.6. Classification of force systems. 1.4 FORCE AND COMPONENTS A single force F acting on a particle may be replaced by two or more forces that, together, have the same effect on the particle. These forces are called components of the original force F. Each force can be resolved into an infinite number of possible sets of components. Resolution of a Force - a given force is replaced by two components which are equivalent to the given force. The most common two-dimensional resolution of a force vector is into rectangular components. y 𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃𝑥 Fx 𝐹𝑦 = 𝐹𝑠𝑖𝑛𝜃𝑥 Fy Θy F Or Fy 𝐹𝑥 = 𝐹𝑠𝑖𝑛𝜃𝑦 θx x Fx 𝐹𝑦 = 𝐹𝑐𝑜𝑠𝜃𝑦 Figure 1.7. Rectangular components Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 11  Or using the slope: The rectangular components Fx and Fy are considered positive if they are in the positive directions of the x- and y-axes and negative if directed oppositely. Note: the orientation of the x- and y-axes is arbitrary. When the rectangular components of a force are known, they completely specify the magnitude, inclination and direction of a force. F = √𝑭𝒙 𝟐 + 𝑭𝒚 𝟐 |𝑭𝒚 | 𝛉𝒙 = 𝑻𝒂𝒏−𝟏 |𝑭𝒙 | The direction of F is determined by the sign of its components and its inclination by the acute angle it makes with the axis. Examples: 1. Compute the x and y component of each of the four forces shown. y Given: T= 722 lb P = 200 lb 3 2 60o x 35o 1 2 Q = 400 lb F = 448 lb Required: Tx, Ty, Px, Py, Fx, Fy, Qx, and Qy Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 12  Solution: +y T= 722 lb Tx Px P = 200 lb Ty √13 3 Py 2 60o +x 35o √5 Qy 1 Fy 2 Q = 400 lb Qx Fx F = 448 lb 2 𝑇𝑥 = − ( ) (722) = −400.49 𝑙𝑏 √13 3 𝑇𝑦 = ( ) (722) = 600.74 𝑙𝑏 √13 𝑃𝑥 = 200𝑐𝑜𝑠60𝑜 = 100 𝑙𝑏 𝑃𝑦 = 200𝑠𝑖𝑛60𝑜 = 173.21 𝑙𝑏 2 𝐹𝑥 = ( ) (448) = 400.70 𝑙𝑏 √5 1 𝐹𝑦 = − ( ) (448) = −200.35 𝑙𝑏 √5 𝑄𝑥 = −40𝑐𝑜𝑠35𝑜 = −327.66 𝑙𝑏 𝑄𝑦 = −400𝑠𝑖𝑛35𝑜 = −229.43 𝑙𝑏 2. Referring to the figure below, determine the components of force P along the x-y axes which are parallel and perpendicular to the incline. Given: Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 13  Required: Px and Py Solution: 𝜃 =𝛼+𝛽 2 𝛼 = tan−1 ( ) = 33.69𝑜 3 3 P = 361 lb 𝛽 = tan−1 ( ) = 36.87𝑜 4 𝜃 = 33.69𝑜 + 36.87𝑜 = 70.56𝑜 For Px and Py: α Py 𝑃𝑥 = 361cos(70.56𝑜 ) θ β 𝑃𝑥 = 120.15 𝑙𝑏 ↖ Px 𝑃𝑥 = 361sin(70.56𝑜 ) β 𝑃𝑦 = 340.42 𝑙𝑏 ↙ 3. The body on the 30o incline is acted upon by a force P inclined at 20o with the horizontal. If P resolved into components parallel and perpendicular to the incline and the value of the parallel component is 300 lb, compute the value of the perpendicular component and of P. Given: component of P parallel to the incline = 300 lb. Required: The value of the perpendicular component, Pn and of P. Solution: Perpendicular component, Pn: 𝑃 tan 50𝑜 = 300 𝑛 𝑃𝑛 = 357.53 𝑙𝑏 ↘ Pn 30 Magnitude of P: o 300 cos 50𝑜 = 𝑃 𝑃 = 466.72 𝑙𝑏 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 14  Components of Forces in Space The direction of a force F is defined by the coordinates of two points located on its line of action. The rectangular components of a force are directly proportional to the rectangular components of the distance d separating the two points. It is expressed by 𝐹𝑥 𝐹𝑦 𝐹𝑧 𝐹 = = = = 𝐹𝑚 𝑥 𝑦 𝑧 𝑑 y + d F Fx x + x Fy y Fz z + z Figure 1.8. Proportionality of force components to distance components where: Fm = force multiplier (in units of force per unit length) x, y, and z = component of distance d between two points along the line of action of the force. To compute for the distance components, subtract the coordinate of the starting point from the final point (final – initial). The magnitudes of F and d can be obtained by repeated applications of the Pythagorean theorem: 𝐹 = √𝐹𝑥 2 + 𝐹𝑦 2 + 𝐹𝑧 2 and 𝑑 = √𝑥 2 + 𝑦 2 + 𝑦 2 𝐹 = 𝐹𝑚 √𝑥 2 + 𝑦 2 + 𝑦 2 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 15  An alternate method of expressing components of a force is in terms of the angles it makes with the coordinate axes. 𝐹𝑥 = 𝐹 cos 𝜃𝑥 y 𝐹𝑦 = 𝐹 cos 𝜃𝑦 + 𝐹𝑧 = 𝐹 cos 𝜃𝑧 Where θx, θy, and θz are the direction cosines of the force F. 𝐹𝑥 Fy F 𝜃𝑥 = cos−1 𝐹 θy Fy 𝐹𝑦 θx Fx + 𝜃𝑦 = cos−1 𝐹 O x 𝐹𝑧 θz 𝜃𝑧 = cos−1 Fz 𝐹 Fz Use the absolute value of the components in determining the direction cosines. Fx To find the third angle once any two have been + z specified: cos 2 𝜃𝑥 + cos2 𝜃𝑦 + cos 2 𝜃𝑧 = 1 Figure 1.9 Direction cosines of the force Examples: 1. Determine (a) the x, y, and z components of the 450-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. Given: F = 450 N Required: a) Fx, Fy, and Fz; b) θx, θy, and θz Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 16  Solution: 𝐹𝐻 = 𝐹 cos 35𝑜 = 450 cos 35𝑜 𝐹𝐻 = 368.62 𝑁 a) 𝐹𝑥 = −𝐹𝐻 sin 40𝑜 Fy = −368.62 sin 40𝑜 𝐹𝑥 = −236.94 𝑁 FH 40o Fz 𝐹𝑦 = 𝐹 sin 35𝑜 Fx = 450 sin 35𝑜 𝐹𝑦 = 258.11 𝑁 𝐹𝑧 = 𝐹𝐻 cos 40𝑜 = 368.62 cos 40𝑜 𝐹𝑧 = 282.38 𝑁 𝐹𝑥 b) 𝜃𝑥 = cos−1 𝐹 −1 236.94 = cos 450 𝑜 𝜃𝑥 = 58.23 𝐹𝑦 𝜃𝑦 = cos −1 𝐹 258.11 = cos−1 450 𝜃𝑦 = 55𝑜 𝐹𝑧 𝜃𝑧 = cos −1 𝐹 −1 282.38 = cos 450 𝑜 𝜃𝑧 = 51.13 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 17  2. A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. Given: F = 385 N in cable Required: Fx, Fy, and Fz of the force exerted by the cable on the support at D. Solution: FBD of the force on cable D 𝐹𝑥 𝐹𝑦 𝐹𝑧 𝐹 = = = = 𝐹𝑚 𝑥 𝑦 𝑧 𝑑 Solve for the distance components along the line of action F (from D to B): 𝑥 = 𝑥𝐵 − 𝑥𝐷 𝑦 = 𝑦𝐵 − 𝑦𝐷 𝑧 = 𝑧𝐵 − 𝑧𝐷 = 480 − 0 = 0 − 510 = 600 − 280 𝑥 = 480 𝑚𝑚 𝑦 = −510 𝑚𝑚 𝑧 = 320 𝑚𝑚 F Magnitude the distance, d: 𝑑 = √𝑥 2 + 𝑦 2 + 𝑦 2 = √(480)2 + (510)2 + (320)2 𝑑 = 770 𝑚𝑚 Force multiplier, Fm: 𝐹 𝐹𝑚 = 𝑑 385 = 770 𝑁 𝐹𝑚 = 0.50 𝑚𝑚 For the x, y, z components: 𝐹𝑥 = 𝑥𝐹𝑚 𝐹𝑧 = 𝑧𝐹𝑚 = (480)(0.50) = (320)(0.50) 𝐹𝑥 = +240 𝑁 𝐹𝑧 = 160 𝑁 𝐹𝑦 = 𝑦𝐹𝑚 = (−510)(0.50) 𝐹𝑦 = −255 𝑁 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 18  EXPLAIN For more examples, read pages 19-31 of your textbook. ELABORATE After going through the prepared lessons, answer the following questions/statements. 1. State the “principle of transmissibility.” 2. Differentiate the external effect from internal effect of a force. 3. Give three characteristics of a force. 4. If a given force is to be resolved into components, how many sets of components can be obtained? Elaborate your answer. Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 19  UNIT 2: VECTOR ALGEBRA At the end of this unit you should be able to apply vector algebra to solve problems involving forces. ENGAGE Before studying this unit, take the test to determine how much you already know about vector algebra. Direction: Choose the letter of the correct answer. Write your answer on the space provided before each number. 1. Determine the scalar product of A = 6.0i + 4.0j - 2.0k and B = 5.0i - 6.0j - 3.0k. A. 30i + 24j + 6k B. 30i - 24j + 6k C. 12 D. 60 2. Determine the angle between the directions of vector A = 3.00i + 1.00j and vector B = -3.00i + 3.00j. A. 26.6° B. 30.0° C. 88.1° D. 117° 3. What is the vector product of A = 2.00i + 3.00j + 1.00k and B = 1.00i - 3.00j - 2.00k? A. -3.00i + 5.00j - 9.00k B. -5.00i + 2.00j - 6.00k C. -9.00i - 3.00j - 3.00k D. -4.00i + 3.00j - 1.00k 4. What is the magnitude of the cross product of a vector of magnitude 2.00 m pointing east and a vector of magnitude 4.00 m pointing 30.0° west of north? A. 4.00 B. -4.00 C. 6.93 D. -6.93 5. If C = -4i - 2j - 3k, its magnitude is A. 5 B. 5.39 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 21  C. 9 D. 9.23 True or False. On the space provided before each number, write True if the statement is true, write False if the statement is false. 6. If all the components of a vector are equal to 1, then that vector is a unit vector. 7. If the dot product of two nonzero vectors is zero, the vectors must be perpendicular to each other. 8. If two nonzero vectors point in the same direction, their dot product must be zero. 9. If two vectors are perpendicular to each other, their cross product must be zero. 10. If two vectors point in opposite directions, their cross product must be zero. Refer to the Answer Key and check your answers. What is your score? ______/10 Answer Key: 10. True 9. False 8. False 7. True 6. False B 5. C 4. A 3. D 2. C 1. EXPLORE Read and understand the prepared lessons and find out how much you have learned about vector algebra by answering the activities under the Elaborate and Evaluation sections. 1.5 VECTOR NOTATION The distinction between magnitude and direction is achieved by a notation which expresses a vector as the product of its magnitude and a unit vector which defines its direction. Vectors – mathematical quantities possessing magnitude and direction. A vector may be:  Fixed or bound vector - vector used to represent a force acting on a given particle has a well-defined point of application.  Sliding vector - forces acting on a rigid body are represented by vectors that can be moved along their lines of action.  Free vector - one which is independent of the point of application of the vector. Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 22  Vector are denoted by printing it in boldface (F), or drawing a short arrow above the letter used to represent it (𝐹⃗ )or by underlining the letter (F). The magnitude of the vector |F| may be denoted by or F. 1.6 UNIT VECTORS A unit vector is defined as a vector of unit magnitude in a specified direction. Multiplying a unit vector by a scalar denotes a vector having the direction of the unit vector and a magnitude equal to that of the scalar. y Let: î = unit vector directed along the x-axis + Ĵ = unit vector directed along the y-axis k̂ = unit vector directed along the z-axis d A ĵ n̂ = a unit vector in an arbitrary direction n̂ such as ⃗⃗⃗⃗⃗⃗ 𝑂𝐴 in Figure 1.10. O î x + x k̂ y Thus, force F having the rectangular coordinates z Fx, Fy, and Fz may be written in the standard + Cartesian form of representing a vector: z Figure 1.10. Unit vectors ̂ 𝐅 = 𝐹𝑥 𝐢̂ + 𝐹𝑦 𝐣̂ + 𝐹𝑧 𝐤 The unit vector n̂ which characterizes the direction of the force F can be obtained by dividing the vector d by the magnitude d of the distance of the two points along its line of action. ̂ 𝐝 𝑥𝐢̂ + 𝑦𝐣̂ + 𝑧𝐤 n̂ = = 𝑑 𝑑 Where: 𝑑 = √𝑥 2 + 𝑦 2 + 𝑧 2 The force F may be expressed in the following forms: 𝐹 𝐅 = 𝐹n̂ = (𝑥î + 𝑦ĵ + 𝑧k̂) = 𝐹𝑚 (𝑥î + 𝑦ĵ + 𝑧k̂) 𝑑 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 23  1.7 VECTOR ALGEBRA A. Dot Product P The dot product is also known as the scalar product and a dot between two vectors is used to denote their multiplication. The dot product of two vectors P and Q, Figure 1.11, is defined as the product of their magnitudes times the cosine of the angle θ between them. θ It is written as Q 𝐏 ∙ 𝐐 = 𝑃𝑄𝑐𝑜𝑠𝜃 Figure 1.11. Two vectors P and Q Using components: Note that: î ∙ î = Ĵ ∙ Ĵ = k̂ ∙ k̂ = 1 î ∙ Ĵ = Ĵ ∙ k̂ = k̂ ∙ î = 0 and 𝐏 ∙ 𝐐 = 𝑃𝑥 𝑄𝑥 + 𝑃𝑦 𝑄𝑦 + 𝑃𝑧 𝑄𝑧 The dot product is a scalar quantity, it may be positive, negative, or zero. If two nonzero vectors are perpendicular their dot product is zero. Properties of the dot product: Commutative: P∙Q=Q∙P Associative: mP ∙ nQ = mnP ∙ Q Distributive F ∙ (P + Q) = F ∙ P + F ∙ Q Applications of the dot product: 1. Angle formed by two given vectors. If 𝐏 = 𝑃𝑥 î + 𝑃𝑦 ĵ + 𝑃𝑧 k̂ and 𝐐 = 𝑄𝑥 î + 𝑄𝑦 ĵ + 𝑄𝑧 k̂ then 𝑃𝑄𝑐𝑜𝑠𝜃 = 𝑃𝑥 𝑄𝑥 + 𝑃𝑦 𝑄𝑦 + 𝑃𝑧 𝑄𝑧 P 𝑃𝑥 𝑄𝑥 + 𝑃𝑦 𝑄𝑦 + 𝑃𝑧 𝑄𝑧 𝑐𝑜𝑠𝜃 = 𝑃𝑄 2. Projection of a vector on a given axis Rule: The component of a vector in any direction is θ L the dot product of the vector with a unit vector in Q the desired direction. O 𝐏∙𝐐 𝑃 cos 𝜃 = = n̂𝑄 ∙ 𝐏 𝑄 Figure 1.12. The projection P and a vector Q along OL Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 24  Examples: 1. In the system shown, it is found that the force y multiplier of force F acting from B to D is Fm = 150 lb/ft and that of force P acting from A to E is Pm = 100 lb/ft. Find the component of each force along AC. What angle does each forcemake with AC? z Given: x F acting from B to D, Fm = 150 lb/ft P acting from A to E, Pm = 100 lb/ft Required: FAC, PAC, θF, and θP Solution: y a) For the components of F and P along AC. 𝐅∙𝐀𝐂 𝐏∙𝐀𝐂 𝐹𝐴𝐶 = 𝐴𝐶 = n̂𝐴𝐶 ∙ 𝐅 and 𝑃𝐴𝐶 = 𝐴𝐶 = n̂𝐴𝐶 ∙ 𝐏 P For F: F = 𝐹𝑚 (𝑥î + 𝑦ĵ + 𝑧k̂) z F x = 0 – 8 = - 8 ft x Force Facts from B to D, y = 0 – (- 3) = 3 ft i.e, x = xD - xB, y = yD - yB, and z = 6 – 0 = 6 ft z = zD - zB F = 150(−8î + 3ĵ + 6k̂) For P: P = 𝑃𝑚 (𝑥î + 𝑦ĵ + 𝑧k̂) x = 0 – 12 = - 12 ft Force Pacts from A to E, y = 4 – 0 = 4 ft i.e, x = xE – xA, y = yE – yA, and z = - 6 – 0 = - 6 ft z = zE – zA P = 100(−12î + 4ĵ − 6k̂) For n̂𝐴𝐶 : 1 n̂𝐴𝐶 = 𝐴𝐶 (𝑥î + 𝑦ĵ + 𝑧k̂) x = 0 – 12 = - 12 ft For the direction of AC, y = - 9 – 0 = - 9 ft i.e, x = xC – xA, y = yC – yA, and z = 0 – 0 = 0 ft z = zC – zA Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 25  𝐴𝐶 = √(−12)2 + (−9)2 + (0)2 = 15 𝑓𝑡 1 ̂𝐴𝐶 = 𝐧 15 (−12î − 9ĵ + 0k̂) For 𝐹𝐴𝐶 : 𝐹𝐴𝐶 = n̂𝐴𝐶 ∙ 𝐅 1 = 15 (−12î − 9ĵ + 0k̂) ∙ 150(−8î + 3ĵ + 6k̂) 150 = [(−12)(−8) + (−9)(3) + (0)(6)] 15 𝐹𝐴𝐶 = 690 𝑙𝑏 For PAC: 𝑃𝐴𝐶 = n̂𝐴𝐶 ∙ 𝐏 1 = (−12î − 9ĵ + 0k̂) ∙ 100(−12î + 4ĵ − 6k̂) 15 100 = [(−12)(−12) + (−9)(4) + (0)(−6)] 15 𝑃𝐴𝐶 = 720 𝑙𝑏 b) For the angle each force make with AC: 𝑃𝐴𝐶 = 𝑃 cos 𝜃𝑃 𝐹𝐴𝐶 = 𝐹 cos 𝜃𝐹 𝑃𝐴𝐶 𝜃𝑃 = cos −1 𝑃 𝐹𝐴𝐶 𝜃𝐹 = cos−1 𝐹 𝑃 = 𝑃𝑚 √𝑥 2 + 𝑦 2 + 𝑦 2 𝐹 = 𝐹𝑚 √𝑥 2 + 𝑦 2 + 𝑦 2 = 100√(−12)2 + (4)2 + (−6)2 = 150√(−8)2 + (3)2 + (6)2 𝐹 = 1400 𝑙𝑏 𝐹 = 150√109 𝑙𝑏 720 𝜃𝑃 = cos −1 1400 690 𝜃𝐹 = cos−1 150 √109 𝜃𝐹 = 59.05𝑜 𝜃𝐹 = 63.86𝑜 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 26  B. Cross Product The cross product is also known as the vector product and the symbol “⨯” between two vectors is used to denote their multiplication. The cross product of two vectors P and Q, Figure 1.13, is defined as the product of their magnitudes times the sine of the angle θ between them. It is written as P⨯Q 𝐏 ⨯ 𝐐 = 𝑃𝑄𝑠𝑖𝑛𝜃 Q There are always two directions perpendicular to a given plane, one on each side of the plane. θ The direction of 𝐏 ⨯ 𝐐 is perpendicular to the plane of the two vectors being multiplied, as given by the right-hand rule. P Figure 1.13. P⨯Q is perpendicular to The right-hand rule: plane of P and Q Point fingers of right hand along the direction P and curl fingers toward Q. The thumb points in direction of P⨯Q. Note that the vector product of two parallel or antiparallel vectors is always zero. In particular, the vector product of any vector with itself is zero. 𝐢̂⨯ 𝐢̂= 0 ̂ 𝐢̂ ⨯ 𝐣̂= 𝐤 ̂ ⨯ 𝐣̂= -𝐢̂ 𝐤 𝐣̂⨯ 𝐣̂ = 0 ̂ 𝐣̂ ⨯ 𝐢̂ = - 𝐤 ̂ = -𝐣̂ 𝐢̂ ⨯ 𝐤 ̂ ⨯𝐤 𝐤 ̂ =0 ̂ = 𝐢̂ 𝐣̂ ⨯ 𝐤 ̂ ⨯ 𝐢̂ = 𝐣̂ 𝐤 Using components: 𝐏 ⨯ 𝐐 = (Pxî+ Pyĵ+ Pzk̂) ⨯ (Qxî+ Qyĵ+ Qzk̂) ̂ 𝐏 ⨯ 𝐐 = (𝑷𝒚 𝑸𝒛 − 𝑷𝒛 𝑸𝒚 ) 𝐢̂ + (𝑷𝒛 𝑸𝒙 − 𝑷𝒙 𝑸𝒛) 𝐣̂ + (𝑷𝒙 𝑸𝒚 − 𝑷𝒚 𝑸𝒛) 𝐤 The vector product can also be expressed in determinant form: 𝐢̂ 𝐣̂ ̂ 𝐤 𝐏 ⨯ 𝐐 = | 𝑃𝑥 𝑃𝑦 𝑃𝑧 | 𝑄𝑥 𝑄𝑦 𝑄𝑧 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 27  Properties of the cross product: Not commutative: P⨯Q≠Q⨯P But: P⨯Q=−Q⨯P Associative when multiplied by a scalar: m(P ⨯ Q) = mP ⨯ Q = P ⨯ mQ Not associative when multiplied by a vector: (F ⨯ P) ⨯ Q ≠ (F ⨯ P) ⨯ Q Distributive F ⨯ (P + Q) = F ⨯ P + F ⨯ Q Applications of the cross product: 1. Shortest distance between a point to a line. 2. Length of the common perpendicular between two nonintersecting vectors. 3. Moment of a force about a point. Examples: 1. Given the vectors a = 2i – 3j – 4k, b = 4i + 2j + k, and c = 3i – j – 2k, evaluate a) a ⨯ b b) c ⨯ a c) (a ⨯ b) ⨯ (a ⨯ c) Solution: 𝑎𝑥 𝑎𝑦 𝑎𝑧 𝑎𝑥 𝑎𝑦 a) 𝐚 ⨯ 𝐛 = |𝑏𝑥 𝑏𝑦 𝑏𝑧 | 𝑏𝑥 𝑏𝑦 The products of terms lying on a diagonal 𝐢 𝐣 𝐤 𝐢 𝐣 downward to the right (↘) are positive (+), 2 −3 −4 2 −3 whereas the products of terms on a diagonal = |4 2 1| 4 2 lying upward to the right (↗) are negative (−). 𝐢 𝐣 𝐤 𝐢 𝐣 = [(−3)(1) − (2)(−4)]𝐢 + [(−4)(4) − (1)(2)]𝐣 + [(2)(2) − (4)(−3)]𝐤 𝐚 ⨯ 𝐛 = 5𝐢 − 18𝐣 + 16𝐤 𝑐𝑥 𝑐𝑦 𝑐𝑧 𝑐𝑥 𝑐𝑦 𝑎 b) 𝐜 ⨯ 𝐚 = | 𝑥 𝑎𝑦 𝑎𝑧 | 𝑎𝑥 𝑎𝑦 𝐢 𝐣 𝐤 𝐢 𝐣 3 −1 −2 3 −1 = |2 −3 −4| 2 −3 𝐢 𝐣 𝐤 𝐢 𝐣 = [(−1)(−4) − (−3)(−2)]𝐢 + [(−2)(2) − (−4)(3)]𝐣 + [(3)(−3) − (2)(−1)]𝐤 𝐜 ⨯ 𝐛 = −2𝐢 + 8𝐣 − 7𝐤 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 28  c) (𝐚 ⨯ 𝐛) ⨯ (𝐚 ⨯ 𝐜) For 𝐚 ⨯ 𝐜: 𝐚 ⨯ 𝐜 = −𝐜 ⨯ 𝐚 = −(−2 + 8𝐣 − 7𝐤) 𝐚 ⨯ 𝐜 = 2𝐢 − 8𝐣 + 7𝐤 5 −18 16 5 −18 (𝐚 ⨯ 𝐛) ⨯ (𝐚 ⨯ 𝐜) = |2 −8 7 | 2 −8 𝐢 𝐣 𝐤 𝐢 𝐣 = [(−18)(7) − (−8)(16)]𝐢 + [(16)(2) − (7)(5)]𝐣 + [(5)(−8) − (2)(−18)]𝐤 (𝐚 ⨯ 𝐛) ⨯ (𝐚 ⨯ 𝐜) = 2𝐢 − 3𝐣 − 4𝐤 2. Refer to the cantilever framework shown in the figure and find the shortest distance from point B to line AC and to line AD. Given: 2’ D 8’ 4’ C 8’ A 6’ W 4’ 3’ B Required: shortest distance from point B to the line AC and to the line AD (0,4,-2) Solution: The shortest distance between a point and a line is the perpendicular distance. (0,4,8) Let: dAC = shortest distance from point B to the line AC. α (8,0,0) dAD = shortest distance from dAD θ point B to the line AD. dAC (3,-6,4) Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 29  For dAC: From the right Δ formed by lines AB, AC and dAC: −8 4 8 −8 4 𝐀𝐂 ⨯ 𝐀𝐁 = | −5 −6 4 | −5 −6 𝑑𝐴𝐶 = 𝐴𝐵 sin 𝜃 𝐢 𝐣 𝐤 𝐢 𝐣 𝐀𝐂 ⨯ 𝐀𝐁 = (𝐴𝐶)(𝐴𝐵) sin 𝜃 = [(4)(4) − (−6)(8)]𝐢 +[(8)(−5) − (4)(−8)]𝐣 |𝐀𝐂⨯𝐀𝐁| 𝑑𝐴𝐶 𝑑𝐴𝐶 = +[(−8)(−6) − (−5)(4)]𝐤 𝐴𝐶 𝐀𝐂 = 𝑥𝐢 + 𝑦𝐣 + 𝑧𝐤 𝐀𝐂 ⨯ 𝐀𝐁 = 64𝐢 − 8𝐣 + 68𝐤 𝐀𝐂 = −8𝐢 + 4𝐣 + 8𝐤 |𝐀𝐂 ⨯ 𝐀𝐁| = √(64)2 + (8)2 + (68)2 𝐴𝐶 = √(8)2 + (4)2 + (8)2 |𝐀𝐂 ⨯ 𝐀𝐁| = √8784 |𝐀𝐂⨯𝐀𝐁| √8784 𝐴𝐶 = 12 𝑓𝑡 𝑑𝐴𝐶 = = 𝐴𝐶 12 𝐀𝐁 = −5𝐢 − 6𝐣 + 4𝐤 𝑑𝐴𝐶 = 7.81 𝑓𝑡 For dAD: From the right Δ formed by lines AB, AD and dAD: −8 4 −2 −8 4 𝐀𝐃 ⨯ 𝐀𝐁 = |−5 −6 4 | −5 −6 𝑑𝐴𝐷 = 𝐴𝐵 sin 𝜃 𝐢 𝐣 𝐤 𝐢 𝐣 𝐀𝐃 ⨯ 𝐀𝐁 = (𝐴𝐷)(𝐴𝐵) sin 𝛼 = [(16) − (12)]𝐢 + [(10) − (−32)]𝐣 + [(48) − (−20)]𝐤 𝑑𝐴𝐷 |𝐀𝐃⨯𝐀𝐁| 𝐀𝐃 ⨯ 𝐀𝐁 = 4𝐢 + 42𝐣 + 68𝐤 𝑑𝐴𝐷 = 𝐴𝐷 |𝐀𝐃 ⨯ 𝐀𝐁| = √(4)2 + (42)2 + (68)2 𝐀𝐃 = 𝑥𝐢 + 𝑦𝐣 + 𝑧𝐤 |𝐀𝐃 ⨯ 𝐀𝐁| = √6404 𝐀𝐃 = −8𝐢 + 4𝐣 − 2𝐤 |𝐀𝐃⨯𝐀𝐁| √6404 𝐴𝐷 = √(8)2 + (4)2 + (2)2 𝑑𝐴𝐷 = 𝐴𝐷 = √84 𝐴𝐶 = √84 𝑓𝑡 𝑑𝐴𝐷 = 8.73 𝑓𝑡 𝐀𝐁 = −5𝐢 − 6𝐣 + 4𝐤 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 30  3. In the figure, if the force multiplier of a force P acting from A to D is Pm = 20 lb/ft, determine the component of P that is perpendicular to the plane defined by points E, A, and C. Given: P acting from A to D Pm = 20 lb/ft Required: component of P that is perpendicular to the plane defined by points E, A, and C, P⫠EAC Solution: (0,3,6) (0,0,-3) P (8,0,0) (4,-5,0) (0,-10,0) Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 31  A direction perpendicular to P = 𝑃𝑚 (𝑥𝐢 + 𝑦𝐣 + 𝑧𝐤) the plane defined by points E, P = 20(−8𝐢 + 0𝐣 − 3𝐤) A, and C can be determined by getting the direction of the Let: n = unit vector perpendicular to plane EAC. cross product of any two sides Then: 𝑃⫠𝐸𝐴𝐶 = 𝐏 ∙ 𝐧 of the plane. For n: e.g. EA ⨯ EC 𝐝 𝐀𝐄 ⨯ 𝐀𝐂 CA ⨯ CE 𝐧 = 𝑑 = |𝐀𝐄 ⨯ 𝐀𝐂| AE ⨯ AC AE = −8𝐢 + 3𝐣 − 6𝐤 AC = −8𝐢 − 10𝐣 + 0𝐤 −8 3 6 −8 3 𝐀𝐄 ⨯ 𝐀𝐂 = |−8 −10 0| −8 −10 𝐢 𝐣 𝐤 𝐢 𝐣 = [(0) − (−60)]𝐢 + [(−48) − (0)]𝐣 + [(80) − (−24)]𝐤 𝐀𝐄 ⨯ 𝐀𝐂 = 60𝐢 − 48𝐣 + 104𝐤 |𝐀𝐄 ⨯ 𝐀𝐂| = √(60)2 + (48)2 + (104)2 |𝐀𝐄 ⨯ 𝐀𝐂| = √16720 𝑓𝑡 2 𝐀𝐄 ⨯ 𝐀𝐂 Therefore: 𝐧 = |𝐀𝐄 ⨯ 𝐀𝐂| 60𝐢−48𝐣+104𝐤 𝐧= √16720 𝑃⫠𝐸𝐴𝐶 = 𝐏 ∙ 𝐧 60𝐢−48𝐣+104𝐤 𝑃⫠𝐸𝐴𝐶 = 20(−8𝐢 + 0𝐣 − 3𝐤) ∙ √16720 20 = [(−8)(60) + (0)(−48) + (−3)(104)] √16720 𝑃⫠𝐸𝐴𝐶 = −122.5 𝑙𝑏 Note: the negative sign indicates that the direction of the component is in the opposite direction of the cross product of 𝐀𝐄 ⨯ 𝐀𝐂. Hence, the component 𝑃⫠𝐸𝐴𝐶 is in the direction of the cross product of 𝐀𝐂 ⨯ 𝐀𝐄. 𝑃⫠𝐸𝐴𝐶 = 122.5 𝑙𝑏 in the direction of the cross product of 𝐀𝐂 ⨯ 𝐀𝐄. Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 32  4. Refer to the system shown and find the length of the common perpendicular between lines BE and AD. Given: Required: length of the common perpendicular between lines BE and AD. Solution: Let: N = BE ⨯ AD -------> represents the direction of the common perpendicular between lines BE and AD The projection on N of any vector joining BE and AD will be the length of the common perpendicular. Vectors joining BE and AD are AB and DE. n = the unit vector in the direction of N. 𝐍 𝐁𝐄 ⨯ 𝐀𝐃 𝐧 = = |BE 𝑁 ⨯ AD| d = length of the common perpendicular 𝐁𝐄 ⨯ 𝐀𝐃 𝑑 = 𝐀𝐁 ∙ 𝐧 = 𝐀𝐁 ∙ |BE ⨯ AD| BE = −4𝐢 + 8𝐣 + 6𝐤 AD = −8𝐢 + 0𝐣 − 3𝐤 AB = −4𝐢 − 5𝐣 + 0𝐤 −4 8 6 −4 8 𝐁𝐄 ⨯ 𝐀𝐃 = |−8 −0 −3| −8 0 𝐢 𝐣 𝐤 𝐢 𝐣 = [(−24) − (0)]𝐢 + [(−48) − (12)]𝐣 + [(0) − (−64)]𝐤 𝐁𝐄 ⨯ 𝐀𝐃 = −24𝐢 − 60𝐣 + 64𝐤 |𝐁𝐄 ⨯ 𝐀𝐃| = √(24)2 + (60)2 + (64)2 |𝐁𝐄 ⨯ 𝐀𝐃| = √8272 𝑓𝑡 2 Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 33  𝐁𝐄 ⨯ 𝐀𝐃 𝑑 = 𝐀𝐁 ∙ |BE ⨯ AD| −24𝐢−60𝐣+64𝐤 = (−4𝐢 − 5𝐣 + 0𝐤) ∙ √8272 1 = [(−4)(−24) + (−5)(−60) + (0)(64)] √8272 𝑑 = 4.35 𝑓𝑡 1.8 MOMENT OF A FORCE The moment of a force about a point – the product of the magnitude of the force by the perpendicular distance from the point to the line of action of the force. Moment arm – the perpendicular distance from the point to the line of action of the force. Figure 1.14. Example of the concept of moment. The magnitude of the moment of a force F O about a center O is expressed by 𝑀𝑂 = 𝐹𝑑 The magnitude of the moment, MO, measures Figure 1.15. Magnitude of the the tendency of the force F to make the rigid moment of a force. body rotate about a fixed axis directed along MO. In some two-dimensional and many of the three-dimensional problems, it is convenient to use a vector approach for moment O calculations. The moment of F about point O may be represented by the cross-product expression 𝐌𝑂 = 𝐫 ⨯ 𝐅 Where: r = any position vector extending from the moment center to any point on the action line of the force. Figure 1.16. The direction of MO is indicated by the right-hand rule. Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 34  1.9 PRINCIPLE OF MOMENTS (VARIGNON’S THEOREM) The principle of moment states that the moment of a force about any point is equal to the sum of the moments of the components of the force about the same point. Consider force F in Figure 1.17, with components F1 and F2. And since F = F1 + F2, the moment of force F about point O is 𝐌𝑂 = 𝐫 ⨯ 𝐅 = 𝐫 ⨯ (𝐅𝟏 + 𝐅𝟐 ) 𝐌𝑂 = 𝐫 ⨯ 𝐅𝟏 + 𝐫 ⨯ 𝐅𝟐 A general symbolic statement of the theorem is 𝑀𝑂𝑅 = ∑ 𝑀𝑂 = ∑ 𝐫 ⨯ 𝐅 (vector method) Figure 1.17. Force F with For coplanar forces, the principle of moments can components F1 and F2 be used by resolving the force into its rectangular components and then determine the moment using a scalar analysis. Thus, +↻ 𝑀𝑂 = 𝑦𝐹𝑥 − 𝑥𝐹𝑦 (scalar method) In computing moments in coplanar force systems, the scalar method is preferred. Sign Convention When dealing with moments it is important to Figure 1.18. Coplanar force F apply a CONSISTENT sign convention with the direction of the moment. with components Fx and Fy It doesn’t matter in which Look at which way the moment is turning. Is it direction you apply the sign turning CLOCKWISE or COUNTERCLOCKWISE? convention as long as you remain Units of moment of a force CONSISTENT. For instance: If a moment is rotating In the SI system of units, where a force is CLOCKWISE (↻) it is considered a expressed in newtons (N) and a distance in POSITIVE moment. meters (m), the moment of a force is expressed in newton-meters (N-m). If a moment is rotating COUNTERCLOCKWISE it is In the U.S. customary system of units, where a considered a NEGATIVE (↺) force is expressed in pounds and a distance in moment. feet or inches, the moment of a force is expressed in lb-ft or lb-in. Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without the prior written permission of SLU, is strictly prohibited. 35  Coplanar Applications Examples: 1. In the figure given, assuming clockwise moments as positive, compute the moment of force F = 450 lb and of force P = 361 lb about points A, B,C and D. Given: F = 450 lb and P = 361 lb, clockwise moments are positive Required: 𝑀𝐴𝐹 , 𝑀𝐵𝐹 , 𝑀𝐶𝐹 , 𝑀𝐷𝐹 and 𝑀𝐴𝑃 , 𝑀𝐵𝑃 , 𝑀𝐶𝑃 , 𝑀𝐷𝑃 Fx Solution: The slope of the line of action of the forces can be determined based on the squares. Fy Resolve the force into their components: 4 𝐹𝑥 = (5) (450) = 360 𝑙𝑏 Py 3 𝐹𝑦 = (5) (450) = 270 𝑙𝑏 2 Px 𝑃𝑥 = ( ) (361) = 200.25 𝑙𝑏 √13 3 𝑃𝑦 = ( ) (361) = 300.37 𝑙𝑏 √13 𝐹 𝐹 𝑀𝐴𝐹 = 𝑀𝐴𝑥 + 𝑀𝐴𝑦 𝑀𝐴𝑃 = 0 (notice the line of action of force P = 0 + (−270)(5) passes through point A) 𝑀𝐴𝐹 = −1350 𝑙𝑏 − 𝑓𝑡 𝑃 𝑀𝐵𝑃 = 𝑀𝐵𝑥 + 𝑀𝐵𝑦 𝑃 𝑀𝐴𝐹 = 1350 𝑙𝑏 − 𝑓𝑡 counterclockwise = (0) + (−300.37)(1) 𝐹 𝐹 𝑀𝐵𝐹 = 𝑀𝐵𝑥 + 𝑀𝐵𝑦 𝑀𝐵𝑃 = −300.37 𝑙𝑏 − 𝑓𝑡 = (360)(6) + (0) 𝑀𝐵𝑃 = 300.37 𝑙𝑏 − 𝑓𝑡counterclockwise 𝑀𝐵𝐹 = 2160 𝑙𝑏 − 𝑓𝑡 clockwise 𝑃 𝑀𝐶𝑃 = 𝑀𝐶𝑥 + 𝑀𝐶 𝑦 𝑃 𝐹 𝐹 𝑀𝐶𝐹 = 𝑀𝐶𝑥 + 𝑀𝐶𝑦 = (−200.25)(3) + (−300.37)(2) = (360)(3) + (270)(1) 𝑀𝐶𝑃 = −1201.49 𝑙𝑏 − 𝑓𝑡 𝑀𝐶𝐹 = 1350 𝑙𝑏 − 𝑓𝑡clockwise 𝑀𝐶𝑃 = 1201.49 𝑙𝑏 − 𝑓𝑡 counterclockwise 𝐹 𝐹 𝑃 𝑃 𝑀𝐷𝐹 = 𝑀𝐷𝑥 + 𝑀𝐷𝑦 𝑀𝐷𝑃 = 𝑀𝐷𝑥 + 𝑀𝐷𝑦 = (360)(6) + (−270)(5) = (0) + (300.37)(4) 𝑀𝐶𝐹 = 810 𝑙𝑏 − 𝑓𝑡clockwise 𝑀𝐷𝑃 = 1201.49 𝑙𝑏 − 𝑓𝑡 clockwise Property of and for the exclusive use of SLU. Reproduction, storing in a retrieval system, distributing, uploading or posting online, or transmitting in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise of any part of this document, without