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ENGR 140: Engineering Mechanics - Statics

ENGR 140: Engineering Mechanics - Statics. Dr. N. Norman School of Mathematics, Science & Engineering Burnette 158, (804)523-5587 Nnorman@Reynolds.edu. Chapter 3: Equilibrium of a Particle. 3.1: Conditions for the Equilibrium of a Particle 3.2: The Free-Body Diagram

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ENGR 140: Engineering Mechanics - Statics

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  1. ENGR 140: Engineering Mechanics -Statics Dr. N. Norman School of Mathematics, Science & Engineering Burnette 158, (804)523-5587 Nnorman@Reynolds.edu

  2. Chapter 3: Equilibrium of a Particle • 3.1: Conditions for the Equilibrium of a Particle • 3.2: The Free-Body Diagram • 3.3: Coplanar Force Systems • 3.4: Three-Dimensional Force Systems

  3. Chapter 3 Objectives More practice with free the body diagram (FBD) Solve particle equilibrium problems using the equations of equilibrium

  4. 3.1: Conditions for the Equilibrium of a Particle • A particle is said to be in equilibrium if • It remains at rest after originally being at rest • It has a constant velocity if originally in motion • Note: Equilibrium in reference to what we are currently doing is “Static Equilibrium” • It satisfies Newton’s first law of motion • The resultant force (the sum of all of the forces) acting on the particle equals zero

  5. 3.2 The Free-Body Diagram Springs • If a linearly undeformedelastic spring is used to support a particle, the length of the spring will change in direct proportion to the force acting on it • A characteristic that defines the elasticity of a spring • Spring constant or stiffness k • The magnitude of force on the said spring is deformed (elongated or compressed) a distance s measured from it unloaded position s = l – lo Spring Force = spring constant * deformation F = k * s

  6. 3.2 The Free-Body Diagram Cables and Pulleys • Unless otherwise stated (in our class) all cables will have negligible weight and cannot be stretched • A cable can only support a tension or pulling force and this force always acts in the direction of the cable

  7. A 30 40 100 lb F1 F2 A A) B) 30 40° 100 lb A F2 F F1 D) C) 30° 40° 30° A A 100 lb 100 lb Think about it… Select the correct FBD of particle A. D

  8. Think About it… When a particle is in equilibrium, the sum of forces acting on it equals ___ . • A constant • A positive number • Zero • A negative number • An integer C

  9. 3.3 Coplanar Force Systems If a particle is subjected to a system of coplanar forces that lie in the x-y plane • Each force can be resolved into its i and j components • For equilibrium these forces produce a zero force resultant (F= 0 ) ( Fx)i+ ( Fy)j+ ( Fz)k= 0

  10. 3.3 Coplanar Force Systems This vector equation will be satisfied when Fx= 0 Fy = 0 Fz = 0 These equations are the three scalar equations of equilibrium. They are valid for any point in equilibrium and allow you to solve for up to three unknowns.

  11. 3.3 Coplanar Force Systems • Note: When solving equations • Two equations can be solved for at most two unknowns • When applying the equations of equilibrium • Account for the sense of direction of any component by using the algebraic sign which corresponds to the arrowhead direction of the component along the axis • When the magnitude is unknown it is okay to assume then change the sign once the sense is known

  12. Think about it… Particle P is in equilibrium with five forces acting on it in 3-D space. How many scalar equations of equilibrium can be written for point P? A) 2 B) 3 C) 4 D) 5 E) 6 B

  13. Think about it… In 3-D, when a particle is in equilibrium, which of the following equations apply? A) ( Fx) i + ( Fy)j+ ( Fz) k= 0 B)  F= 0 C)  Fx =  Fy=  Fz= 0 D) All of the above. E) None of the above. D

  14. 3.3 Coplanar Force SystemsProcedure for Analysis • Equations of equilibrium • Apply the equation of equilibrium • Components are positive it they are directed along a positive axis • If more than two unknowns exist and the problem involves a spring, apply F = ks • The magnitude of a force is always positive • If the solution yields a negative result, the sense of direction is opposite of what is indicated on the FBD

  15. Example Given: The four forces and geometry shown. Find: The forceF5 required to keep particle O in equilibrium. Plan: 1) Draw a FBD of particle O 2) Write the unknown force as F5= {Fxi + Fyj + Fzk} N 3) Write F1, F2 , F3 , F4, F5in Cartesian vector form 4) Apply the 3 equilibrium equations to solve for the 3 unknowns Fx, Fy, and Fz.

  16. Example Continued F1= {300(4/5) j + 300 (3/5) k} N F1= {240 j+ 180 k} N F2= {– 600 i} N F3= {– 900 k} N F4 = F4 (rB/ rB) = 200 N [(3i– 4 j+ 6 k)/(32 + 42 + 62)½] = {76.8 i– 102.4 j+ 153.6 k} N F5 = { Fxi–Fyj+ Fzk} N

  17. Example Continued Equating the i,j, k components to zero, we have ∑Fx= 76.8 – 600 + Fx= 0 ; gives Fx= 523.2 N ∑Fy= 240 –102.4 + Fy= 0 ; gives Fy= – 137.6 N ∑Fz= 180 – 900 + 153.6 + Fz= 0 ; gives Fz= 566.4 N Thus,F5= {523 i–138 j+ 566 k} N Using this force vector, you can determine the force’s magnitude and coordinate direction angles

  18. z F3 = 10 lb P F2 = 10 lb y A F1 = 20 lb x Think about it… Four forces act at point A and point A is in equilibrium. Select the correct force vector P. A) {-20 i+ 10 j– 10 k}lb B) {-10 i– 20 j– 10 k} lb C) {+ 20 i– 10j– 10 k}lb D) None of the above. D

  19. Example Given: A 3500 lb motor and plate, as shown, are in equilibrium and supported by three cables and d = 4 ft. Find: Magnitude of the tension in each of the cables. Plan: Draw a free body diagram of Point A. Let the unknown force magnitudes be FB, FC, F D . 2) Represent each force in Cartesian vector form 3) Apply equilibrium equations to solve for unknowns.

  20. FBD of Point A

  21. Example Continued FBD of Point A z W W= load or weight of unit = 3500 klbFB= FB(rAB/rAB) = FB {(4 i– 3 j– 10 k) / (11.2)} lb FC = FC (rAC/rAC) = FC { (3 j– 10 k) / (10.4 ) }lb FD = FD( rAD/rAD) = FD { (– 4i + 1j–10 k) / (10.8) }lb y x FD FB FC

  22. Example Continued Why can’t you simply add all of the vectors together? The particle A is in equilibrium, hence FB + FC + FD + W = 0 Now equate the respective i, j, k components to zero (i.e., apply the three scalar equations of equilibrium)  Fx= (4/ 11.2)FB– (4/ 10.8)FD= 0  Fy= (– 3/ 11.2)FB+ (3/ 10.4)FC + (1/ 10.8)FD = 0  Fz= (– 10/ 11.2)FB– (10/ 10.4)FC– (10/ 10.8)FD + 3500 = 0 Solving the three simultaneous equations gives FB= 1467 lb, FC= 914 lb, FD= 1420 lb

  23. Chapter 4.1-4.4 Force center resultants

  24. Chapter 4 Objectives: • To discuss moment of force • Show how to calculate it in 2D and 3D • To understand a method for finding the moment of a force about a specified axis • To define the moment of a couple • To discuss methods for determining the resultant of non-concurrent force systems • To lean how to reduce a simple distributed loading to a resultant force have a specified location

  25. 4.1 moment of a force-scalar formation • The moment Mo of a forceF about an axis passing through a specific point O provides a measure of the tendency of the force to cause the body to rotate about the axis (torque). • The moment is a vector it has both magnitude and direction • The magnitude of the moment is determined from Mo = Fd • Where d is the perpendicular (shortest distance) form point O to the line of action of the force F

  26. If a system of forces lies in the x–y plane, then the moment produced by each force about point O will be directed along the z-axis. The resultant moment MRO of the system can be determined by simply adding the moments of all the forces algebraically since all the moment vectors are collinear i.e. + MRO = ∑ Fd 4.1 moment of a force-scalar formation

  27. Beams are often used to bridge gaps in walls. We have to know what the effect of the force on the beam will have on the supports of the beam. APPLICATIONS What do you think is happening at points A and B?

  28. Carpenters often use a hammer in this way to pull a stubborn nail. Through what sort of action does the force FH at the handle pull the nail? How can you mathematically model the effect of force FH at point O? APPLICATIONS (continued)

  29. 4.1 moment of a force-scalar formation • In a 2-D case, the magnitude of the moment is Mo = F d As shown, d is the perpendicular distance from point O to the line of action of the force. In 2-D, the direction of MO is either clockwise (CW) or counter-clockwise (CCW), depending on the tendency for rotation.

  30. F a For example, MO = F d and the direction is counter-clockwise. b O d F F y F x a b O 4.1 moment of a force-scalar formation Often it is easier to determine MO by using the components of F as shown. Then MO = (FY a) – (FX b). 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.

  31. 4.2 Cross product • While finding the moment of a force in 2-D is easy when you know distance d, finding the perpendicular distances can be hard, especially when you are working with forces in 3D. • A general approach to finding the moment of a force exists. • Used when dealing with 3D forces but can be used in the 2D case as well. • Called the cross product of two vectors.

  32. 4.2 Cross product In general, the cross product of two vectors A and Bresults in another vector, C, i.e., C =A  B. The magnitude and direction of the resulting vector can be written as C = A  B= A B sin uC As shown, uCis the unit vector perpendicular to both A and B vectors (or to the plane containing the A and B vectors).

  33. 4.2 Cross product The right-hand rule is a useful tool for determining the direction of the vector resulting from a cross product. For example: i j = k Note that a vector crossed into itself is zero, e.g.,i  i = 0

  34. Also, the cross product can be written as a determinant. Each component can be determined using 2  2 determinants. 4.2 Cross product

  35. 4.3 MOMENT OF A FORCE – VECTOR FORMULATION Moments in 3-D can be calculated using scalar (2-D) approach, but it can be difficult and time consuming. Thus, it is often easier to use a mathematical approach called the vector cross product. Using the vector cross product,MO = r  F. Here r is the position vector from point O to any point on the line of action of F.

  36. 4.3 MOMENT OF A FORCE – VECTOR FORMULATION So, using the cross product, a moment can be expressed as By expanding the above equation using 2  2 determinants (see Section 4.2), we get (sample units are N - m or lb - ft) MO = (ry FZ - rZFy) i  (rxFz - rzFx ) j + (rxFy - ryFx) k The physical meaning of the above equation becomes evident by considering the force components separately and using a 2-D formulation.

  37. 4.3 moment of a force • Principle of Transmissibility • Since r can extend from O to any point on the line of action of F, F is a sliding vector and can act at any point along its line of action and create the same moment about point O. • Resultant Moment of a System of Forces • If a body is acted upon by a system of n forces, the resultant moment about O is just the vector sum of the individual moments. MRO =

  38. LAWS OF OPERATION

  39. CARTESIAN VECTOR FORMULATION • To find the cross product of any two Cartesian vectors A and B we use the determinant • The following useful results can be obtained by applying the right-hand rule and don’t need to be memorized.

  40. 4.4 principles of moments • The principle of moments (Varignon’s theorem) states that the moment of a force about a point is equal to the sum of the moments of the force’s components about the point. • This is particularly convenient since it is often easier to determine the moments of a force’s components rather than the moment of the force itself (e.g., in two dimensions).

  41. F = 12 N d = 3 m • A Think about it… What is the moment of the 12 N force about point A (MA)? A) 3 N·m B) 36 N·m C) 12 N·m D) (12/3) N·mE) 7 N·m B

  42. Think about it… The moment of force F about point O is defined as MO = ___________ . A)r x F B) F x r C) r • F D) r * F A

  43. Given:A 100 N force is applied to the frame. Find: The moment of the force at point O. Plan: Example 1 1) Resolve the 100 N force along x and y-axes. 2) Determine MO using a scalar analysis for the two force components and then add those two moments together..

  44. Solution +  Fy = – 100 (3/5) N +  Fx = 100 (4/5) N + MO = {– 100 (3/5)N (5 m) – (100)(4/5)N (2 m)} N·m = – 460 N·m or 460 N·m CW Example 1

  45. Example 2 Given:F1={100 i - 120 j+ 75 k}lb F2={-200 i+250 j+ 100 k}lb Find:Resultant moment by the forces about point O. Plan: o 1) Find F=F1+F2and rOA 2) Determine MO = rOA F

  46. Example 2 Solution: First, find the resultant force vector F F =F1+F2 = { (100 - 200) i + (-120 + 250) j+ (75 + 100) k} lb = {-100 i+130 j+175 k} lb Find the position vector rOA rOA= {4 i + 5 j+3 k} ft Then find the moment by using the vector cross product. j i k MO = = [{5(175) – 3(130)} i – {4(175) – 3(-100)}j + {4(130) – 5(-100)} k] ft·lb = {485 i– 1000 j + 1020 k} ft·lb 4 5 3 -100 130 175

  47. Chapter 4.5-4.9 Force center resultants

  48. 4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS Recall: when the moment of a force is computed about a point, the moment and its axis are always perpendicular to the plane containing the force and the moment arm. In some problems it is important to find the component of this moment about a specified axis that passes through the point.

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