1. Chapter 1 Forces: Maintaining Equilibrium or Changing Motion

2. Force Force: a push or pull acting on a body that causes or tends to cause a change in the linear motion of the body Characteristics of a force magnitude direction point of application. line of action Net Force: resultant force (overall effect of multiple forces acting on a body) Example: push from side and front = at angle

3. Force Free body diagram - sketch that shows a defined system in isolation with all the force vectors acting on the system.

4. Why is this impossible?

5. Classifying Forces Internal Force: acts within the object or system whose motion is being investigated action / reaction forces both act on different parts of the system tensile-internal pulling forces when the structure is under tension compressive- internal pushing (squeezing) forces act on the ends of an internal structure do not accelerate the body Orientate segments, maintain structural integrity

6. Internal Forces Examples Contraction of muscles Do not accelerate the body

7. Classifying Forces External Force: acts on object as a result of interaction with the environment surrounding it non-contact - occur even if objects are not touching each other gravity, magnetic contact - occur between objects in contact fluid (air & water resistance) reaction forces with another body (ground, implement) vertical (normal) reaction force acts perpendicular to bodies in contact shear reaction force acts parallel to surfaces in contact (friction)

9. F = ma Force may also be defined as the product of a body's mass and the acceleration of that body resulting from the application of the force. Units of force are units of mass multiplied by units of acceleration.

10. Units of Force Metric system (systeme internationale -SI) Newton (N) the amount of force necessary to accelerate a mass of 1 kg at 1 m/s 2 English system pound (lb) the amount of force necessary to accelerate a mass of 1 slug at 1 ft/s 2 equal to 4.45 N

11. Weight (external force) Weight - the amount of gravitational force exerted on a body. wt=ma g. Since weight is a force, units of weight are units of force - either N or lb. As the mass of a body increases, its weight increases proportionally.

12. Weight The factor of proportionality is the acceleration of gravity, which is -9.81m/s 2 or - 32 ft/s 2. The negative sign indicates that the acceleration of gravity is directed downward or toward the center of the earth.

13. Weight On the moon or another planet with a different gravitational acceleration, a body's weight would be different, although its mass would remain the same. Space station example

14. Weight Because weight is a force, it is also characterized by magnitude, direction, and point of application. The direction in which weight acts is always toward the center of the earth.

15. Center of Weight (Gravity) The point at which weight is assumed to act on a body is the body's center of gravity.

16. Friction (external force) Component of a contact force that acts parallel to the surface in contact acts opposite to motion or motion tendency reflects interaction between molecules in contact reflects force “squeezing” surfaces together acts at the area of contact between two surfaces

17. Static friction: surfaces not moving relative to each other Maximum static friction: maximum amount of friction that can be generated between two static surfaces Dynamic friction: surfaces move relative to each other constant magnitude friction during motion always less than maximum static friction Friction

18. F =  N Friction depends on Nature of materials in contact (  ) Force squeezing bodies together (N) Known as the normal contact (reaction) force Friction is FUN

19. Nature of materials in contact Coefficient of friction (  ) value serves as an index of the interaction between two surfaces in contact. Coefficient of Friction (  )

21. Factors Affecting Friction The greater the coefficient of friction, the greater the friction The greater the normal contact force, the greater the friction

22. Normal reaction force (NRF) AKA - normal contact force force acting perpendicular to two surfaces in contact. magnitude intentionally altered to increase or decrease the amount of friction present in a particular situation football coach on sled push (pull) upward to slide object pivot turn on ball of foot Reaction Force

23. Friction Manipulation Examples of manipulating  and N c shoe design

24. Friction Manipulation Examples of manipulating  and N c shoe design grips (gloves, tape, sprays, chalk) skiing: decrease for speed, increase for safety curling your examples???

25. Curling

26. Bode Miller, Silver Medalist 2002

27. Shark Skin Suits in Swimming

28. Friction and Surface Area Friction force is proportional to the normal contact force Friction is not affected by the size of the surface area in contact normal contact force distributed over the area in contact Friction is affected by the nature of the materials in contact

29. Friction and Surface Area With dry friction, the amount of surface area in contact does not impact the amount of friction. Same force acting over more area Same force acting over smaller area

30. Coefficient of Friction Hot, soft, rough surfaces have higher coefficients of friction Tires, concrete, etc Cold, hard, smooth surfaces have lower coefficients of friction Ice, marble, etc

31. Friction Calculate the force of friction when you slide on ice. Given = Normal force of 1000N C of F = 0.02

32. Solution Calculate the force of friction when you slide on ice. Given = Normal force of 1000N C of F = 0.02 F = µ * N F = (0.02) * (1000N) F = 20 N

33. Force Force: a push or pull acting on a body that causes or tends to cause a change in the linear motion of the body (an acceleration of the body) Characteristics of a force magnitude direction point of application line of action sense (push or pull along the line of action) Vector represented with an arrow

34. Recall Concept of Net External Force Must add all forces acting on an object together

35. Free body diagram Free body diagram - sketch that shows a defined system in isolation with all the force vectors acting on the system defined system: the body of interest vector: arrow to represent a force length: size of the force tip: indicates direction location: point of application

36. Free body diagram Me, at rest in front of class sagittal plane view What are the names of the forces? How big are the forces? What direction are the forces? Where are the forces applied? Note: link contains practice problems 

37. Free body diagram Free body diagram ==>static analysis Me (mass= 75 kg): at rest (a = 0) in front of class (assume gravity at -10m/s/s) sagittal plane view Wt & vGRF C of M, at feet 750N, ????? - & + F = m a F = 0 Wt + vGRF = 0 vGRF = - Wt vGRF = - (-750) vGRF = 750 N Note: link contains practice problems

38. Addition of Forces (calculating the net [resultant] force) Net force = vector sum of all external forces acting on the object (body) account for magnitude and direction e.g., Add force of 100N and 200N

39. Addition of Forces (calculating the net [resultant] force) Net force = vector sum of all external forces acting on the object (body) account for magnitude and direction e.g., Add force of 100N and 200N act in same direction??? Act in opposite direction??? Act orthogonal to each other??? Act at angles to each other???

40. Colinear forces Forces have the same line of action May act in same or different directions ie tug of war teammates: 100N, 200N, 400N show force on rope graphically

41. Colinear forces Forces have the same line of action May act in same or different directions ie tug of war teammates: 100N, 200N, 400N tug of war opponents: 200N, 200N, 200N show force on rope graphically

42. Colinear forces Forces have the same line of action May act in same or different directions ie tug of war teammates: 100N, 200N, 400N tug of war opponents: 200N, 200N, 200N calculate resultant of the two teams show force on rope graphically calculate algebraically

43. Free body diagram ==>static analysis Weightlifter (mass 80 kg) 100 kg bar overhead at rest (a = 0) sagittal plane view What are the forces? Where are the forces applied? How big are the forces? What direction are the forces? Free body diagrams a)Weightlifter b)bar Fig 1.19, p 42

44. Free body diagram ==>static analysis Weightlifter (80 kg) 100 kg bar overhead at rest (a = 0) sagittal plane view What are the forces? Where are the forces applied? How big are the forces? What direction are the forces? Wt & vGRF CofM, at feet 800N, ????? - & +

45. Free body diagram ==>static analysis Weightlifter (80 kg) 100 kg bar overhead at rest (a = 0) sagittal plane view Wt, bar & vGRF CofM, on hands, at feet 800N, 1000N, ????? -, -, + F = m a F = 0 Wt + bar + vGRF = 0 vGRF = - Wt - bar vGRF = - (-800) - (- 1000) vGRF = + 1800 N

46. Concurrent Forces Forces do not act along same line, but do act through the same point ie gymnast jumps up to grab bar. Coach stops swinging by applying force to front and back of torso. 20 N posterior directed push on front of torso 30 N anterior directed push on back of torso 550N force from bar on gymnast’s hands gymnast mass 50 kg

47. Concurrent Forces gymnast hanging from grab bar. Coach applies force to front and back of torso to stop swing. 20 N posterior directed push on front of torso 30 N anterior directed push on back of torso 550N force from bar on gymnast’s hands gymnast mass 50 kg Page 29 in book

48. Concurrent Forces gymnast hanging from grab bar. Coach applies force to front and back of torso to stop swing. 20 N posterior directed push on front of torso 30 N anterior directed push on back of torso 550N force from bar on gymnast’s hands gymnast mass 50 kg What is the resultant force? Tip to tail method (fig 1.8 & 1.9 in text) separate algebraic summation of horizontal and negative forces Pythagorean theorem to solve resultant magnitude Inverse tangent to solve direction (angle)

49. Horizontal forces: 20N – 30N = -10N Vertical forces: -500N + 550N = 50N Resultant force: a 2 + b 2 = c 2 (-10N) 2 + (50N) 2 = c 2 (100N 2 ) + (2500N 2 ) = c 2 2600N 2 = c 2 C = 51N

50. Quantifying Kinetics Vector composition - process of determining a single vector from 2 or more vectors through vector addition. Resultant - single vector that results from vector composition.

51. Quantifying Kinetics Vector resolution - breaking down a resultant vector into its horizontal and vertical components. Graphic method. Trigonometric method. Graphic Method

52. Trigonometry: SOH, CAH, TOA Key Equations: sin  = opp/hyp cos  = adj/hyp tan  = opp/adj Pythagoras theorem a 2 + b 2 = c 2 hyp (c) adj opp  Check this site.

53. Vector Composition Sample:  = 30 degrees adj = 100 N Find opp and hyp hyp (c) adj opp 

54. Vector Composition use cos  = adj/hyp tan  = opp/adj hyp (c) adj opp 

55. Vector Composition cos  = adj/hyp cos 30 = 100/hyp hyp = 100/.866 hyp = 115.47 N tan  = opp/adj tan 30 = opp/100 opp =.5774 x 100 opp = 57.74 N hyp (c) adj opp 

56. Vector Resolution Sample  = 35 degrees hyp = 120 N Find opp and adj hyp (c) adj opp 

57. Trigonometric Calculations Use sin  = opp/hyp cos  = adj/hyp hyp (c) adj opp 

58. Vector Resolution sin  = opp/hyp sin 35 = opp/120 opp = 120 X.5736 opp = 68.83 N cos  = adj/hyp cos 35 = adj/120 adj = 120 X.8192 adj = 98.30 N hyp (c) adj opp 

59. Michelle Kwan Note orientation of stance leg.

60. Michelle Kwan Note orientation of stance leg.

61. Michelle Kwan Note orientation of stance leg.

62. Ground Reaction Force Foot Contact

63. Ground Reaction Force Toe Off

64. Resolution of Forces Forces are not colinear and not Hor & Vert? ie figure 1.1: forces on shot 100 N from shot-putters hand mass of shot = 4 kg What is the resultant force on shot? Draw Components of 100N force solve graphically: tedious & imprecise trigonometric technique 100N wt

65. Resolution of Forces Forces are not colinear and not concurrent ie figure 1.1: forces on shot 100 N from shot-putters hand mass of shot = 4 kg W = mg = (4kg)(-10m/s/s)= -40 N What is the resultant force on shot? Draw Components of 100N force solve graphically: tedious & imprecise trigonometric technique 100N wt -40N 60 o

66. Solution Given Hyp = 100N Angle = 60 degrees Sin angle = opp/hyp Sin 60 = opp/hyp Sin 60*hyp = opp (0.866)*(100N)= 86.6N Given Hyp = 100N Angle = 60 deg Cos angle = adj/hyp Cos 60 = adj/hyp Cos 60*hyp = adj (0.500)*(100N) = 50N

67. Free body diagram ==>static analysis Child on swing (20 kg) Mother’s force 40 N horizontal 10 N upward at rest (a = 0) force of swing on child? Fig 1.2 p 43

68. Free body diagram ==>static analysis Child on swing (mass = 20 kg) Mother’s force 40 N horizontal 10 N upward at rest (a = 0) force of swing on child? F x = m a x F y = m a y

69. Solution  F x = R x + 40N = 0 R x = -40N  F y = R y + 10N + (-200N) = 0 R y = 190N