Vector Mechanics for Engineers: Statics Lecture Slides PDF

Ninth Edition VECTOR MECHANICS FOR ENGINEERS: 2 CHAPTER STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Statics of Particles © 2010 The McGraw-Hill Companies, Inc. All rights re...

Ninth Edition VECTOR MECHANICS FOR ENGINEERS: 2 CHAPTER STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Statics of Particles © 2010 The McGraw-Hill Companies, Inc. All rights reserved. NOTES- Thursday, January 16, 2025 (Lecture #2) Change to the syllabus: ◦ April 22nd- ex day, no instruction that day. Might move the Cables lecture to the next week Lecture formats: ◦ Overall, the class will be very interactive, when solving math-based problems in class, the Professor will call on us to answer questions during the lecture. ◦ Not all slide content will be covered during the lecture, instead it might be given as homework or as readings. If we still have questions, join the breakout rooms during the next lecture. ◦ Almost all of our problem solving is in 2D, not many 3D problems in the course. Quizzes/Exams: ◦ Professor: we will not test any vocab de nitions on quizzes/test- on these we'll only solve math problems basically. NO CHEAT SHEETS ! ◦ Study approach: focus on the problems in the slides mainly, then HW, then quizzes, and then move onto textbook problems. Can ask about in breakout rooms or o ce hours. HW grading based on both correctness and completion. inion  Vector Mechanics for Engineers: Statics Edition Ninth Contents Introduction Sample Problem 2.3 Resultant of Two Forces Equilibrium of a Particle Vectors Free-Body Diagrams Addition of Vectors Sample Problem 2.4 Resultant of Several Concurrent Sample Problem 2.6 Forces Rectangular Components in Space Sample Problem 2.1 Sample Problem 2.7 Sample Problem 2.2 Rectangular Components of a Force: Unit Vectors Addition of Forces by Summing Components © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2-2  Vector Mechanics for Engineers: Statics Edition Ninth Objectives How force is split into an x and y Describe force as a vector quantity. component, the resultant, FBDs.. Examine vector operations useful for the analysis of forces. Determine the resultant of multiple forces acting on a particle. Resolve forces into components. Add forces that have been resolved into rectangular components. Introduce the concept of the free-body diagram. Use free-body diagrams to assist in the analysis of planar and spatial particle equilibrium problems. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2-3  Vector Mechanics for Engineers: Statics Edition Ninth Introduction The objective for the current chapter is to investigate the effects of forces on particles: - replacing multiple forces acting on a particle with a single equivalent or resultant force, - relations between forces acting on a particle that is in a state of equilibrium. The focus on particles does not imply a restriction to miniscule bodies. Rather, the study is restricted to analyses in which the size and shape of the bodies is not significant so that all forces may be assumed to be applied at a single point. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2-4  Vector Mechanics for Engineers: Statics Edition Ninth Resultant of Two Forces Force: action of one body on another; characterized by its point of application, magnitude, line of action, and sense. Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force. The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs. Force is a vector quantity. N, kN, lb, kip © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2-5  Vector Mechanics for Engineers: Statics Edition Ninth Vectors Vector: parameters possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations. Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature Vector classifications: - Fixed or bound vectors have well defined points of application that cannot be changed without affecting an analysis. - Free vectors may be freely moved in space without changing their effect on an analysis. - Sliding vectors may be applied anywhere along their line of action without affecting an analysis. Equal vectors have the same magnitude and direction. Negative vector of a given vector has the same magnitude and the opposite direction. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2-6  Vector Mechanics for Engineers: Statics Edition Ninth Addition of Vectors Trapezoid rule for vector addition Triangle rule for vector addition Law of cosines, C B R 2 = P 2 + Q 2 − 2 PQ cos B    C R = P+Q Law of sines, sin A sin B sin C = = B Q R A Vector addition is commutative,     P+Q = Q+ P Vector subtraction © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2-7  Vector Mechanics for Engineers: Statics Edition Ninth Fundamental Properties of Vectors The Parallelogram Law for Addition and the Triangle Law An equivalent statement of the parallelogram law is the triangle law, which is demonstrated below. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2-8  Vector Mechanics for Engineers: Statics Edition Ninth © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2-9  Vector Mechanics for Engineers: Statics Edition Ninth © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 10  Vector Mechanics for Engineers: Statics Edition Ninth Addition of Vectors Addition of three or more vectors through repeated application of the triangle rule The polygon rule for the addition of three or more vectors. Vector addition is associative,          P + Q + S = (P + Q ) + S = P + (Q + S ) Multiplication of a vector by a scalar © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 11  Vector Mechanics for Engineers: Statics Edition Ninth Fundamental Properties of Vectors The subtraction of two vectors A and B, is defined as A – B = A +(-B) as shown in the figure below. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 12  Vector Mechanics for Engineers: Statics Edition Ninth Resultant of Several Concurrent Forces Concurrent forces: set of forces which all pass through the same point. A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces. Vector force components: two or more force vectors which, together, have the same effect as a single force vector. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 13  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.1 SOLUTION: Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the the diagonal. The two forces act on a bolt at A. Determine their resultant. Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 14  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.1 Graphical solution - A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured, R = 98 N  = 35 Graphical solution - A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured, R = 98 N  = 35 Sample Problem 2.1.ipt © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 15  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.1 Trigonometric solution - Apply the triangle rule. From the Law of Cosines, R 2 = P 2 + Q 2 − 2 PQ cos B = (40 N )2 + (60 N )2 − 2(40 N )(60 N ) cos 155 R = 97.73N From the Law of Sines, sin A sin B = Q R Q sin A = sin B R 60 N = sin 155 97.73N A = 15.04  = 20 + A  = 35.04 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 16  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.2 STRATEGY: Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge A barge is pulled by two axis and length proportional to 5000 lb. tugboats. If the resultant of the forces exerted by the Find a trigonometric solution by tugboats is 5000 lbf directed applying the Triangle Rule for vector along the axis of the barge, addition. With the magnitude and determine the tension in each direction of the resultant known and of the ropes for  = 45o. the directions of the other two sides parallel to the ropes given, apply the Discuss with a neighbor how Law of Sines to find the rope tensions. you would solve this problem. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 17  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.2 MODELING and ANALYSIS: Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides. T1 = 3700lb T2 = 2600lb Trigonometric solution - Triangle Rule with Law of Sines T1 T2 5000lb = = sin45 sin30 sin105 T1 = 3660lb T2 = 2590lb © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 18  Vector Mechanics for Engineers: Statics Edition Ninth What if…? At what value of  would the tension in rope 2 be a minimum? Hint: Use the triangle rule and think about how changing  changes the magnitude of T2. After considering this, discuss your ideas with a neighbor. The minimum tension in rope 2 occurs when T1 and T2 are perpendicular. T2 = (5000lb)sin30 T2 = 2500lb T1 = (5000lb)cos30 T1 = 4330lb  = 90 − 30  = 60 REFLECT and THINK: Part (a) is a straightforward application of resolving a vector into components. The key to part (b) is recognizing that the minimum value of T2 occurs when T1 and T2 are perpendicular. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 19  Vector Mechanics for Engineers: Statics Edition Ninth Rectangular Components of a Force: Unit Vectors It’s possible to resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle. Fx and Fy are referred to as rectangular vector components and    F = Fx + Fy   Define perpendicular unit vectors i and j which are parallel to the x and y axes. Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components.    F = Fx i + Fy j  Fx and Fy are referred to as the scalar components of F © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 20  Vector Mechanics for Engineers: Statics Edition Ninth Addition of Forces by Summing X and Y Components To find the resultant of 3 (or more) concurrent forces,     R = P+Q+ S Resolve each force into rectangular components, then add the components in each direction:         R x i + R y j = Px i + Py j + Qx i + Q y j + S x i + S y j = ( Px + Qx + S x )i + (Py + Q y + S y ) j   The scalar components of the resultant vector are equal to the sum of the corresponding scalar components of the given forces. Rx = Px + Qx + S x R y = Py + Q y + S y =  Fx =  Fy To find the resultant magnitude and direction, 2 2 −1 R y R = Rx + R y  = tan Rx © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 21  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.3 STRATEGY: Resolve each force into rectangular components. Determine the components of the resultant by adding the corresponding force components in the x and y directions. Four forces act on bolt A as shown. Calculate the magnitude and direction Determine the resultant of the force of the resultant. on the bolt. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 22  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.3 MODELING: ANALYSIS: Resolve each force into rectangular components. force mag x − comp y − comp F1 150 +129.9 +75.0 F2 80 −27.4 +75.2 F3 110 0 −110.0 F4 100 +96.6 −25.9 R x = +199.1 R y = +14.3 Determine the components of the resultant by adding the corresponding force components. REFLECT and THINK: Calculate the magnitude and direction. Arranging data in a table not only 2 2 R = 199.6 N helps you keep track of the R = 199.1 + 14.3 calculations, but also makes things 14.3 N tan  =  = 4.1  simpler for using a calculator on 199.1 N similar computations. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 23  Vector Mechanics for Engineers: Statics Edition Ninth Equilibrium of a Particle When the resultant of all forces acting on a particle is zero, the particle is in equilibrium. Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line. Particle acted upon by Particle acted upon by three or more forces: two forces: - graphical solution yields a closed polygon - equal magnitude - algebraic solution - same line of action   R = F = 0 - opposite sense  Fx = 0  Fy = 0 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 24  Vector Mechanics for Engineers: Statics Edition Ninth Free-Body Diagrams and Problem Solving Space Diagram: A sketch showing Free Body Diagram: A sketch showing the physical conditions of the only the forces on the selected particle. problem, usually provided with This must be created by you. the problem statement, or represented by the actual physical situation. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 25  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.4 STRATEGY: Construct a free body diagram for the particle at the junction of the rope and cable. Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle. Apply trigonometric relations to determine the unknown force In a ship-unloading operation, a magnitudes. 3500-lb automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope? © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 26  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.4 ANALYSIS: Apply the conditions for equilibrium and solve for the unknown force magnitudes. Law of Sines: TAB T 3500 lb = AC = sin 120 sin 2 sin 58 MODELING: T AB = 3570 lb T AC = 144 lb © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 27  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.4 REFLECT and THINK: This is a common problem of knowing one force in a three-force equilibrium problem and calculating the other forces from the given geometry. This basic type of problem will occur often as part of more complicated situations in this text. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 28  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.6 STRATEGY: Decide what the appropriate “body” is and draw a free body diagram The condition for equilibrium states that the sum of forces equals 0, or:   It is desired to determine the drag force R = F = 0 at a given speed on a prototype sailboat  Fx = 0  Fy = 0 hull. A model is placed in a test channel and three cables are used to The two equations means we can solve align its bow on the channel centerline. for, at most, two unknowns. Since For a given speed, the tension is 40 lb there are 4 forces involved (tensions in in cable AB and 60 lb in cable AE. 3 cables and the drag force), it is easier Determine the drag force exerted on the to resolve all forces into components hull and the tension in cable AC. and apply the equilibrium conditions © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 29  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.6 MODELING and ANALYSIS: The correct free body diagram is shown and the unknown angles are: 7 ft 1.5 ft tan  = = 1.75 tan  = = 0.375 4 ft 4 ft  = 60.25  = 20.56 In vector form, the equilibrium condition requires that the resultant force (or the sum of all forces) be zero:      R = T AB + TAC + TAE + FD = 0 Write each force vector above in component form. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 30  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.6 Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions. r r r TAB = −(40 lb)sin60.26 i + (40 lb)cos60.26 j r r = −(34.73 lb)i + (19.84 lb) j r r r TAC = TAC sin20.56 i + TAC cos20.56 j r r = 0.3512TAC i + 0.9363TAC j r r TAE = −(60 lb) j r r FD = FD i r R=0 r = (−34.73+ 0.3512TAC + FD ) i r + (19.84 + 0.9363TAC − 60) j © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 31  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.6  R=0  = (− 34.73 + 0.3512 TAC + FD ) i  + (19.84 + 0.9363TAC − 60) j This equation is satisfied only if each component of the resultant is equal to zero ( Fx = 0) 0 = −34.73+ 0.3512TAC + FD ( Fy = 0) 0 =19.84 + 0.9363TAC − 60 REFLECT and THINK: In TAC = +42.9 lb drawing the free-body diagram, you assumed a sense for each FD = +19.66 lb unknown force. A positive sign in the answer indicates that the assumed sense is correct. You can draw the complete force polygon (above) to check the results. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 32  Vector Mechanics for Engineers: Statics Edition Ninth Expressing a Vector in 3-D Space If angles with some of the axes are given:  Resolve Fh into  Resolve F into The vector F is horizontal and vertical rectangular components contained in the components. Fx = Fh cos plane OBAC. Fy = F cos  y = F sin y cos Fh = F sin  y Fz = Fh sin 𝟐 𝟐 𝟐 = F sin y sin 𝑭© 2010 = The√𝑭 + 𝑭 𝒙 Companies, McGraw-Hill + 𝑭 𝒚 Inc.𝒛All rights reserved. 2 - 33  Vector Mechanics for Engineers: Statics Edition Ninth Expressing a Vector in 3-D Space If the direction cosines are given:  With the angles between F and the axes, Fx = F cos  x Fy = F cos  y Fz = F cos  z     F = Fx i + Fy j + Fz k  = F (cos  x i + cos  y j + cos  z k )    = F      = cos  x i + cos  y j + cos  z k    is a unit vector along the line of action of F and cos  x , cos   y , and cos  z are the direction cosines for F © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 34  Vector Mechanics for Engineers: Statics Edition Ninth Expressing a Vector in 3-D Space If two points on the line of action are given: Direction of the force is defined by the location of two points, M ( x1 , y1 , z1 ) and N ( x2 , y 2 , z 2 ) r d = vector joining M and N r r r = dxi + dy j + dz k d x = x 2 − x1 d y = y 2 − y1 d z = z 2 − z1 r r F = F r r r r = 1 d ( dxi + dy j + dz k) Fd x Fd y Fd z Fx = Fy = Fz = d d © 2010 The McGraw-Hill Companies, Inc. All rights reserved. d 2 - 35  Vector Mechanics for Engineers: Statics Edition Ninth © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 36  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.7 STRATEGY: Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B. Apply the unit vector to determine the components of the force acting on A. Noting that the components of the unit vector are the direction cosines for the The tension in the guy wire is 2500 N. vector, calculate the corresponding Determine: angles. a) components Fx, Fy, Fz of the force acting on the bolt at A, b) the angles x, y, z defining the direction of the force (the direction cosines) © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 37  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.7 MODELING and ANALYSIS: Determine the unit vector pointing from A towards B. r r r AB = (−40m)i + (80m) j + (30m)k (−40m) + (80m) + (30m) 2 2 2 AB = = 94.3 m r  −40  r  80  r  30  r  = i +   j + k  94.3   94.3   94.3   r r r = −0.424 i + 0.848 j + 0.318k Determine the components of the force. r r F = F r r r  ( = (2500 N) −0.424 i + 0.848 j + 0.318k ) r r r = (−1060 N )i + (2120 N) j + (795 N)k © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 38  Vector Mechanics for Engineers: Statics Edition Ninth Sample Problem 2.7 Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.      = cos  x i + cos  y j + cos  z k    = −0.424 i + 0.848 j + 0.318k  x = 115.1  y = 32.0  z = 71.5 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 39  Vector Mechanics for Engineers: Statics Edition Ninth What if…? FBA Since the force in the guy wire must be the same throughout its length, the force at B (and acting toward A) must be the same magnitude but opposite in FAB direction to the force at A. r r FBA = −FAB r r r What are the components of the = (1060 N)i + (−2120 N) j + (−795 N)k force in the wire at point B? Can you find it without doing any REFLECT and THINK: It makes sense calculations? that, for a given geometry, only a certain set  of components and angles characterize a Give this some thought and discuss given resultant force. The methods in this this with a neighbor. section allow you to translate back and forth between forces and geometry. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 40