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ENGR 152: Statics Lecture #2

Professor Martinez. ENGR 152: Statics Lecture #2. COMMON CONVERSION FACTORS. Work problems in the units given unless otherwise instructed!. 1 ft = 0.3048 m 1 lb = 4.4482 N 1 slug = 14.5938 kg. Example: Convert a torque value of 47 in • lb into SI units. Answer is 5.31026116 N • m.

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ENGR 152: Statics Lecture #2

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  1. Professor Martinez ENGR 152: StaticsLecture #2

  2. COMMON CONVERSION FACTORS • Work problems in the units given unless otherwiseinstructed! • 1 ft = 0.3048 m • 1 lb = 4.4482 N • 1 slug = 14.5938 kg • Example: Convert a torque value of 47 in • lb into SI units. • Answer is 5.31026116 N • m

  3. Problems • Convert 2 km/h to m/s. How many ft/s is this? • Evaluate each of the following and express with SI units having an appropriate prefix: • (50 mN)(6 GN) • (400 mm)(0.6 MN)² • 45 MN³/900 Gg

  4. THE INTERNATIONAL SYSTEM OF UNITS (Section 1.4) • No Plurals (e.g., m = 5 kg not kgs ) • Separate Units with a • (e.g., meter second = m • s ) • Most symbols are in lowercase. • Some exceptions are N, Pa, M and G. • Exponential powers apply to units, e.g., cm • cm = cm2 • Compound prefixes should not be used. • Other rules are given in your textbook (pg. 10).

  5. NUMERICAL CALCULATIONS (Section 1.5) • Must have dimensional “homogeneity.” Dimensions have to be the same on both sides of the equal sign, (e.g. distance = speed  time.) • Use an appropriate number of significant figures (3 for answer, at least 4 for intermediate calculations). Why? • Be consistent when rounding off. • - greater than 5, round up (3528  3530) • - smaller than 5, round down (0.03521  0.0352) • - equal to 5, round up.

  6. PROBLEM SOLVING STRATEGY: IPE, A 3 Step Approach 1. Interpret:Read carefully and determine what is given and what is to be found/ delivered. Ask, if not clear. If necessary, make assumptions and indicate them. 2. Plan: Think about major steps (or a road map) that you will take to solve a given problem. Think of alternative/creative solutions and choose the best one. 3. Execute: Carry out your steps. Use appropriate diagrams and equations. Estimate your answers. Avoid simple calculation mistakes. Reflect on / revise your work.

  7. ATTENTION Review 1. For a statics problem your calculations show the final answer as 12345.6 N. What will you write as your final answer? A) 12345.6 N B) 12.3456 kN C) 12 kN D) 12.3 kN E) 123 kN 2. In three step IPE approach to problem solving, what does P stand for? A) Position B) Plan C) Problem D) Practical E) Possible

  8. FORCE VECTORS, VECTOR OPERATIONS & ADDITION COPLANAR FORCES Today’s Objective: Students will be able to : a) Resolve a 2-D vector into components. b) Add 2-D vectors using Cartesian vector notations. • In-Class activities: • Reading Review • Application of Adding Forces • Parallelogram Law • Resolution of a Vector Using • Cartesian Vector Notation (CVN) • Addition Using CVN • Attention Review

  9. READING Review 1. Which one of the following is a scalar quantity? A) Force B) Position C) Mass D) Velocity 2. For vector addition you have to use ______ law. A) Newton’s Second B) the arithmetic C) Pascal’s D) the parallelogram

  10. READING QUIZ 1. Which one of the following is a scalar quantity? A) Force B) Position C) Mass D) Velocity 2. For vector addition you have to use ______ law. A) Newton’s Second B) the arithmetic C) Pascal’s D) the parallelogram

  11. APPLICATION OF VECTOR ADDITION There are four concurrent cable forces acting on the bracket. How do you determine the resultant force acting on the bracket ?

  12. SCALARS AND VECTORS (Section 2.1) ScalarsVectors Examples: mass, volume force, velocity Characteristics: It has a magnitude It has a magnitude (positive or negative) and direction Addition rule: Simple arithmetic Parallelogram law Special Notation: None Bold font, a line, an arrow or a “carrot” In the PowerPoint presentation vector quantity is represented Like this (in bold, italics, and yellow).

  13. VECTOR OPERATIONS(Section 2.2) Scalar Multiplication and Division

  14. VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE Parallelogram Law: Triangle method (always ‘tip to tail’): How do you subtract a vector? How can you add more than two concurrent vectors graphically ?

  15. Vector Subtraction • R’ = A – B = A + (-B) • Example, pg. 19

  16. “Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse. RESOLUTION OF A VECTOR

  17. CARTESIAN VECTOR NOTATION (Section 2.4) • We ‘resolve’ vectors into components using the x and y axes system. • Each component of the vector is shown as a magnitude and a direction. • The directions are based on the x and y axes. We use the “unit vectors” i and j to designate the x and y axes.

  18. For example, F = Fx i + Fy j or F' = F'x i + F'y j The x and y axes are always perpendicular to each other. Together,they can be directed at any inclination.

  19. ADDITION OF SEVERAL VECTORS • Step 1 is to resolve each force into its components • Step 3 is to find the magnitude and angle of the resultant vector. • Step 2 is to add all the x components together and add all the y components together. These two totals become the resultant vector.

  20. Example of this process,

  21. You can also represent a 2-D vector with a magnitude and angle.

  22. EXAMPLE Given: Three concurrent forces acting on a bracket. Find: The magnitude and angle of the resultant force. • Plan: • Resolve the forces in their x-y components. • Add the respective components to get the resultant vector. • c) Find magnitude and angle from the resultant components.

  23. EXAMPLE (continued) F1 = { 15 sin 40° i + 15 cos 40° j } kN = { 9.642 i + 11.49 j } kN F2 = { -(12/13)26 i + (5/13)26 j } kN = { -24 i + 10 j } kN F3 = { 36 cos 30° i– 36 sin 30° j } kN = { 31.18 i– 18 j } kN

  24. y FR  x EXAMPLE (continued) Summing up all the i and j components respectively, we get, FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN = { 16.82 i + 3.49 j } kN FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN  = tan-1(3.49/16.82) = 11.7°

  25. GROUP PROBLEM SOLVING Given: Three concurrent forces acting on a bracket Find: The magnitude and angle of the resultant force.

  26. GROUP PROBLEM SOLVING (continued) F1 = { (4/5) 850 i - (3/5) 850 j } N = { 680 i - 510 j } N F2 = { -625 sin(30°) i - 625 cos(30°) j } N = { -312.5 i - 541.3 j } N F3 = { -750 sin(45°) i + 750 cos(45°) j } N { -530.3 i + 530.3 j } N

  27. y x  FR GROUP PROBLEM SOLVING (continued) Summing up all the i and j components respectively, we get, FR = { (680 – 312.5 – 530.3) i + (-510 – 541.3 + 530.3) j }N = { - 162.8 i - 521 j } N • FR = ((162.8)2 + (521)2)½ = 546 N • = tan–1(521/162.8) = 72.64° or From Positive x axis  = 180 + 72.64 = 253 °

  28. y x 30° F = 80 N ATTENTION Review 1. Resolve Falong x and y axes and write it in vector form. F = { ___________ } N A) 80 cos (30°) i - 80 sin (30°) j B) 80 sin (30°) i + 80 cos (30°) j C) 80 sin (30°) i - 80 cos (30°) j D) 80 cos (30°) i + 80 sin (30°) j 2. Determine the magnitude of the resultant (F1 + F2) force in N when F1 = { 10 i + 20 j } N and F2 = { 20 i + 20 j } N . A) 30 N B) 40 N C) 50 N D) 60 N E) 70 N

  29. Homework • Complete Problems: • 1-6 (pg. 15) • 1-8 (pg. 15) • 2-10 (pg. 28)* • 2-32 (pg. 38)* • 2-57 (pg. 42)* • Complete all work in pencil • Grading on 1-5 scale • Looking for thoroughness (don’t just write the answer) • Due next Thursday at beginning of class

  30. End of the Lecture Let Learning Continue

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