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CE 102 Statics

CE 102 Statics. Chapter 2 Statics of Particles. Contents. Introduction Resultant of Two Forces Vectors Addition of Vectors Resultant of Several Concurrent Forces Sample Problem 2.1 Sample Problem 2.2 Rectangular Components of a Force: Unit Vectors

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CE 102 Statics

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  1. CE 102 Statics Chapter 2 Statics of Particles

  2. Contents Introduction Resultant of Two Forces Vectors Addition of Vectors Resultant of Several Concurrent Forces Sample Problem 2.1 Sample Problem 2.2 Rectangular Components of a Force: Unit Vectors Addition of Forces by Summing Components Sample Problem 2.3 Equilibrium of a Particle Free-Body Diagrams Sample Problem 2.4 Sample Problem 2.5 Rectangular Components in Space Sample Problem 2.6

  3. 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.

  4. Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force. Resultant of Two Forces • force: action of one body on another; characterized by its point of application, magnitude, line of action, and sense. • The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs. • Force is a vector quantity.

  5. Vector: parameters possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations. • Equal vectors have the same magnitude and direction. • Negative vector of a given vector has the same magnitude and the opposite direction. Vectors • 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.

  6. Trapezoid rule for vector addition • Triangle rule for vector addition • Law of cosines, C B C • Law of sines, B • Vector addition is commutative, • Vector subtraction Addition of Vectors

  7. 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, • Multiplication of a vector by a scalar Addition of Vectors

  8. 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. Resultant of Several Concurrent Forces

  9. 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.

  10. 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, • 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, Sample Problem 2.1

  11. Sample Problem 2.1 • Trigonometric solution - Apply the triangle rule.From the Law of Cosines, From the Law of Sines,

  12. Sample Problem 2.2 • SOLUTION: • Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge axis and length proportional to 5000 N. A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5000 N directed along the axis of the barge, determine • Find a trigonometric solution by applying the Triangle Rule for vector addition. With the magnitude and direction of the resultant known and the directions of the other two sides parallel to the ropes given, apply the Law of Sines to find the rope tensions. • the tension in each of the ropes for a = 45o, • the value of a for which the tension in rope 2 is a minimum. • The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in a.

  13. Sample Problem 2.2 • Graphical solution - Parallelogram Rule with known resultant direction and magnitude, known directions for sides. • Trigonometric solution - Triangle Rule with Law of Sines

  14. Sample Problem 2.2 • The angle for minimum tension in rope 2 is determined by applying the Triangle Rule and observing the effect of variations in a. • The minimum tension in rope 2 occurs when T1 and T2 are perpendicular.

  15. May resolve a force vector into perpendicular components so that the resulting parallelogram is a rectangle. are referred to as rectangular vector components and • Define perpendicular unit vectors 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.Fx and Fyare referred to as the scalar components of Rectangular Components of a Force: Unit Vectors

  16. Wish to find the resultant of 3 or more concurrent forces, • Resolve each force into rectangular components • The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces. • To find the resultant magnitude and direction, Addition of Forces by Summing Components

  17. Sample Problem 2.3 • SOLUTION: • Resolve each force into rectangular components. • Determine the components of the resultant by adding the corresponding force components. • Calculate the magnitude and direction of the resultant. Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

  18. SOLUTION: • Resolve each force into rectangular components. • Determine the components of the resultant by adding the corresponding force components. • Calculate the magnitude and direction. Sample Problem 2.3

  19. Particle acted upon by three or more forces: • graphical solution yields a closed polygon • algebraic solution 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 two forces: • equal magnitude • same line of action • opposite sense

  20. Free-Body Diagram: A sketch showing only the forces on the selected particle. Free-Body Diagrams Space Diagram: A sketch showing the physical conditions of the problem.

  21. Sample Problem 2.4 • SOLUTION: • Construct a free-body diagram for the particle at the junction of the rope and cable. • Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle. • Apply trigonometric relations to determine the unknown force magnitudes. In a ship-unloading operation, a 3500-N 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?

  22. Sample Problem 2.4 • SOLUTION: • Construct a free-body diagram for the particle at A. • Apply the conditions for equilibrium. • Solve for the unknown force magnitudes.

  23. Sample Problem 2.5 • SOLUTION: • Choosing the hull as the free body, draw a free-body diagram. • Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero. It is desired to determine the drag force at a given speed on a prototype sailboat hull. A model is placed in a test channel and three cables are used to align its bow on the channel centerline. For a given speed, the tension is 40 N in cable AB and 60 N in cable AE. Determine the drag force exerted on the hull and the tension in cable AC. • Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.

  24. Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero. Sample Problem 2.5 • SOLUTION: • Choosing the hull as the free body, draw a free-body diagram.

  25. Sample Problem 2.5 • Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions. r r r ( ) ( ) = - ° + ° T 40 N sin 60 . 26 i 40 N cos 60 . 26 j AB r r ( ) ( ) = - + 34 . 73 N i 19 . 84 N j r r r = ° + ° T T sin 20 . 56 i T cos 20 . 56 j AC AC AC r r = + 0 . 3512 T i 0 . 9363 T j AC AC r r ( ) = - T 6 0 N j r r = F F i D D r = R 0 r ( ) = - + + 34 . 73 0 . 3512 T F i AC D r ( ) + + - 19 . 84 0 . 9363 T 60 j AC

  26. Sample Problem 2.5 This equation is satisfied only if each component of the resultant is equal to zero.

  27. The vector is contained in the plane OBAC. • Resolve into horizontal and vertical components. = f F F cos x h = q f F sin cos y = f F F sin z h = q f F sin sin y Rectangular Components in Space • Resolve into rectangular components.

  28. With the angles between and the axes, • is a unit vector along the line of action ofand are the direction cosines for Rectangular Components in Space

  29. Direction of the force is defined by the location of two points, Rectangular Components in Space

  30. Sample Problem 2.6 • SOLUTION: • Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B. • Apply the unit vector to determine the components of the force acting on A. • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles. The tension in the guy wire is 2500 N. Determine: a) components Fx, Fy, Fz of the force acting on the bolt at A, b) the angles qx, qy, qzdefining the direction of the force

  31. SOLUTION: • Determine the unit vector pointing from A towards B. • Determine the components of the force. Sample Problem 2.6

  32. Sample Problem 2.6 • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

  33. 240 lb A 30o a 75 lb 50o 75 lb Problem 2.7 The direction of the 75-lb forces may vary, but the angle between the forces is always 50o. Determine the value of a for which the resultant of the forces acting at A is directed horizontally to the left.

  34. Problem 2.7 240 lb A 30o a 75 lb 50o 75 lb Solving Problems on Your Own The direction of the 75-lb forces may vary, but the angle between the forces is always 50o. Determine the value of a for which the resultant of the forces acting at A is directed horizontally to the left. 1. Determine the resultant R of two or more forces. 2. Draw a parallelogram with the applied forces as two adjacent sides and the resultant as the included diagonal. 3. Set the resultant, or sum of the forces, directed horizontally.

  35. Problem 2.7 Solution 240 lb A 30o a 75 lb 50o 75 lb Determine the resultant R of two or more forces. We first Replace the two 75-lb forces by their resultant R1, using the triangle rule. a R1 = 2(75 lb) cos25o = 135.95 lb 25o 25o 50o R1 R1 = 135.95 lb a +25o

  36. Problem 2.7 Solution sin(a+25o) sin(30o) = 240 lb 135.95 lb (240 lb) sin(30o) sin(a+25o) = = 0.88270 135.95 lb Draw a parallelogram with the applied forces as two adjacent sides and the resultant as the included diagonal. Set the resultant, or sum of the forces, directed horizontally. R2 Consider the resultant R2 of R1 and the 240-lb force and recall that R2 must be horizontal and directed to the left. a+25o 30o R1 = 135.95 lb 240 lb Law of sines: a + 25o = 61.97o a = 37.0o

  37. y 360 mm C 450 mm D O 500 mm B 320 mm z A x 600 mm Problem 2.8 A container of weight W = 1165 N is supported by three cables as shown. Determine the tension in each cable.

  38. Problem 2.8 y 360 mm C 450 mm D O 500 mm B 320 mm z A x 600 mm F d Solving Problems on Your Own A container of weight W = 1165 N is supported by three cables as shown. Determine the tension in each cable. 1. Draw a free-body diagram of the particle. This diagram shows the particle and all the forces acting on it. 2. Resolve each of the forces into rectangular components. Follow the method outlined in the text. F = F l= (dxi + dyj + dzk)

  39. Problem 2.8 y 360 mm C 450 mm D O 500 mm B 320 mm z A x 600 mm Solving Problems on Your Own A container of weight W = 1165 N is supported by three cables as shown. Determine the tension in each cable. 3. Set the resultant, or sum, of the forces exerted on the particle equal to zero. You will obtain a vectorial equation consisting of terms containing the unit vectors i, j, and k. Three scalar equations result, which can be solved for the unknowns.

  40. Problem 2.8 Solution y 360 mm C 450 mm D O TAC 500 mm B 320 mm TAD TAB z x A 600 mm W =_ (1165 N) j Draw a free-body diagram of the particle. S F = 0 TAB+ TAC + TAD + W = 0 AB = (450 mm)i + (600 mm)j AB = 750 mm AC = (600 mm)j_ (320mm)kAC = 680 mm AD = (_500 mm)i + (600 mm)j + (360 mm)kAD = 860 mm

  41. y Problem 2.8 Solution 360 mm C 450 mm D O AB TAC TAB= TABlAB = TAB = 500 mm AB B 320 mm TAD TAB ( ) 600 z 450 x j A i = = + TAB 600 mm 750 750 W =_ (1165 N) j = (0.6 i + 0.8 j) TAB 320 AC TAC k 680 AC AD 500 600 360 k i TAD + j + 860 860 860 AD 25 30 18 k i + j+ 43 43 43 Resolve each of the forces into rectangular components. ( ) ( ) 600 8 15 k j j = TAC = TAClAC = = TAC TAC _ _ 680 17 17 ( ) TAD = TADlAD = = TAD = ( ) = TAD

  42. y Problem 2.8 Solution 360 mm C 450 mm D O TAC 500 mm B 320 mm TAD TAB z x A 600 mm W =_ (1165 N) j 25 0.6 TAB_ TAD = 0 43 15 30 TAD 17 43 18 8 TAD = 0 43 17 Set the resultant, or sum, of the forces exerted on the particle equal to zero. Substitution into S F = 0, factor i, j, k and set their coefficients to zero: TAB = 0.9690 TAD (1) 0.8 TAB+ TAC + _ 1165 N = 0 + (2) TAC = 0.8895 TAD _ TAC + (3)

  43. y Problem 2.8 Solution 360 mm C 450 mm D O TAC 500 mm B 320 mm TAD TAB z x A 600 mm W =_ (1165 N) j 15 30 17 43 Substitution for TAB and TAC from (1) and (3) into (2): ( 0.8 x 0.9690 + x 0.8895 + )TAD_ 1165 N = 0 TAD = 516 N TAB = 500 N TAC = 459 N 2.2578 TAD_ 1165 N = 0 From (1): TAB = 0.9690 (516 N) From (3): TAC = 0.8895 (516 N)

  44. Problem 2.9 y A Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles qx, qy, and qz defining the direction of that force. 56 ft D a O B 20o 50o z C x

  45. y Problem 2.9 A 56 ft D a O B 20o 50o z C x Solving Problems on Your Own Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles qx, qy, and qz defining the direction of that force. 1. Determine the rectangular components of a force defined by its magnitude and direction. If the direction of the force F is defined by the angles qy and f, projections of F through these angles or their components will yield the components of F.

  46. y Problem 2.9 A 56 ft D a O B 20o 50o z C x Fx Fy Fz cos qx= cos qy= cos qz= F F F Solving Problems on Your Own Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles qx, qy, and qz defining the direction of that force. 2. Determine the direction cosines of the line of action of a force. The direction cosines of the line of action of a force F are determined by dividing the components of the force by F.

  47. y Problem 2.9 Solution From triangle AOB: 56 ft cos qy = = 0.86154 65 ft qy = 30.51o A Determine the direction cosines of the line of action of a force. 65 ft qy F Fy 56 ft Fx B O (a)Fx = _ F sin qy cos 20o = _ (3900 lb) sin 30.51o cos 20o Fx = _1861 lb 20o Fz z x Fy = + F cos qy = (3900 lb)(0.86154) Fy = + 3360 lb Fz = + (3900 lb) sin 30.51o sin 20oFz = + 677 lb

  48. y Problem 2.9 Solution A 65 ft qy Fx _1861 lb F Fy (b) cos qx = = F 3900 lb Fx B O 20o Fz x Fz 677 lb cos qz = = + = + 0.1736 qz = 80.0o F 3900 lb Determine the direction cosines of the line of action of a force. 56 ft cos qx = _ 0.4771 qx = 118.5o From above (a): qy = 30.5o

  49. B A Two cables are tied together at C and loaded as shown. determine the tension (a) in cable AC, (b) in cable BC. 9 ft 8.5 ft C 5 ft 396 lb 12 ft 7.5 ft Problem 2.10

  50. Problem 2.10 B A 9 ft 8.5 ft C 5 ft 396 lb 12 ft 7.5 ft Solving Problems on Your Own Two cables are tied together at C and loaded as shown. determine the tension (a) in cable AC, (b) in cable BC. 1. Draw a free-body diagram of the particle. This diagram shows the particle and all the forces acting on it. 2. Set the resultant, or sum, of the forces exerted on the particle equal to zero. You will obtain a vectorial equation consisting of terms containing the unit vectors i, j, and k. Three scalar equations result, which can be solved for the unknowns.

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