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Fundamentals of Statics and Dynamics - ENGR 3340

Fundamentals of Statics and Dynamics - ENGR 3340. Professor: Dr. Omar E. Meza Castillo omeza@bayamon.inter.edu http://facultad.bayamon.inter.edu/omeza Department of Mechanical Engineering. Tentative Lectures Schedule.

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Fundamentals of Statics and Dynamics - ENGR 3340

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  1. Fundamentals of Statics and Dynamics - ENGR 3340 Professor: Dr. Omar E. Meza Castillo omeza@bayamon.inter.edu http://facultad.bayamon.inter.edu/omeza Department of Mechanical Engineering

  2. Tentative Lectures Schedule

  3. One thing you learn in science is that there is no perfect answer, no perfect measure. A. O. Beckman Topic 8: Moment of a Force Scalar Formulation

  4. Objectives • To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions • To provide a method for finding the moment of a force about a specific axis • To define the moment of a couple • To present methods for determining the resultants of nonconcurrent force systems • To indicate how to reduce a simple distributed loading to a resultant force having a specified location

  5. Moment of a Force. Applications What is the net effect of the two forces on the wheel? What is the effect of the 30 N force on the lug nut?

  6. Moment of a Force. Moment in 2-D The moment of a force about a point provides a measure of the tendency for rotation(sometimes called a torque). Moment about z-axis Moment about x-axis No moment

  7. Moment of a Force.Moment in 2-D – Scalar Formulation • The moment is a scalar or vector? • In the 2-D case, the magnitude of the moment is As shown, d is theperpendicular distance from point O to the line of action of the force. • In 2-D, the direction of MO depends on the tendency for rotation, • clockwise ,or counter-clockwise + + • The resultant moment of a system of coplanar forces is defined as: +

  8. F a b O d Fy a Fx b O Moment of a Force. Moment in 2-D – Scalar Formulation • For example, • magnitude, MO = F d • direction, counter-clockwise + • Often it is easier to determine MO by using the components of F + Note the different signs on the terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate because of the force.

  9. ApplicationProblems

  10. 1. What is the moment of the 10 N force about point A (MA)? A) 3 N·m B) 36 N·m C) 12 N·m D) (12/3) N·m E) 7 N·m F = 12 N d = 3 m • A Quiz

  11. Example 1

  12. Example 1

  13. Example 2

  14. Example 3 • A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O. • Determine: • Moment about O, • Horizontal force at A which creates the same moment, • Smallest force at A which produces the same moment, • Location for a 240-lb vertical force to produce the same moment, • Whether any of the forces from b, c, and d is equivalent to the original force.

  15. Solution 3 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.

  16. Solution 3 Horizontal force at A that produces the same moment,

  17. Solution 3 The smallest force at A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA.

  18. Solution 3 To determine the point of application of a 240 lb force to produce the same moment,

  19. Solution 3 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.

  20. Given: A 20 lb force is applied to the hammer. Find: The moment of the force at A. Example 4 y x Plan: Since this is a 2-D problem: 1) Resolve the 20 lb force along the handle’s x and y axes. 2) Determine MA using a scalar analysis.

  21. Solution 4 y Solution: +  Fy = 20 sin 30° lb +  Fx = 20 cos 30° lb x + MA = {–(20 cos 30°)lb (18 in) – (20 sin 20°)lb (5 in)} = – 351.77 lb·in = 352 lb·in (clockwise)

  22. Homework5  http://facultad. bayamon.inter.edu/omeza/ Omar E. Meza Castillo Ph.D.

  23. ¿Preguntas? Comentarios

  24. GRACIAS

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