1. Forces Maintaining Equilibrium or Changing Motion
Chapter 1
Forces Maintaining Equilibrium or Changing Motion

3. Objectives Define force Classify forces Define friction force
Define weight
Determine the resultant of two or more forces
Resolve a force into component forces acting at right angles to each other

4. Objectives
Determine whether an object is in static equilibrium, if the forces acting on the object are known
Determine an unknown force acting on an object, if all the other forces acting on the object are known and the object is in static equilibrium

5. What Are Forces?
Forces enable us to start moving, stop moving, and change directions
We manipulate the forces acting on us to maintain our balance in stationary positions (e.g. cyclist, diver)
Simply defined, a force is a push or a pull
Forces are exerted by objects on other objects

6. What Are Forces?
Forces come in pairs; the force exerted by one object on another is matched by an equal but oppositely directed force exerted by the second object on the first—Action and reaction
Forces accelerate objects—Start, stop, speed up, slow down, or change direction
SI unit is N (Newton)
1N force is the force that would accelerate a 1kg mass 1m/s2
1N = .225lb
1lb = 4.448N

7. What Are Forces?
Describe the force a shot-putter exerts on a shot at the instant shown in Figure 1.1
Factors we want to know:
Size
Point of application
Direction (line of action)
Sense (pushes or pulls along the line of action)
Force is known as a vector
Mathematical representation of anything described by size or direction

9. Internal Forces
Forces that act within the object or system whose motion is being investigated
Muscles pull on tendons, which pull on bones
At joints, bones push on cartilage
Tensile forces—Pulling forces acting on the ends of an internal structure
Compressive forces—Pushing forces acting on the ends of an internal structure

10. Internal Forces
Sometimes the tensile or compressive forces acting on a structure are greater than the internal forces the structure can withstand—Structure fails or breaks—Muscle pulls, tendon ruptures, ligament tears, bone breaks
Muscles can only produce internal forces—They are incapable of producing changes in the motion of the body without external forces
Body is only able to change its motion if it can push or pull against some external object (e.g. ground reaction force)

12. External Forces
Forces that act on an object as a result of its interaction with the environment
Non-Contact forces—Objects not touching (e.g. gravity, magnetic, electrical)—Gravity is primarily studied in sports biomechanics
Gravity (g) accelerates objects downward at 9.81m/s2 no matter how large or small the object, and any place on earth (ignoring air resistance)

13. External Forces
Weight—In biomechanics, force of gravity acting on an object—Don’t confuse with weight as commonly thought of being in lbs from a bathroom scale
W = mg
W = weight (measured in N)
m = mass (measured in kg)
g = acceleration due to gravity
How much do you weigh?

14. External Forces
Contact forces—Occur between objects in contact with each other
Objects in contact can be solid or fluid (e.g. air resistance and water resistance are fluid)
Forces between solid objects studied primarily in sports biomechanics—An athlete and some other object (opponent, ground, implement)
To put a shot, an athlete must be holding the shot
To jump up, an athlete must be in contact with the ground and push down
To run forward, an athlete must push backward against ground

15. External Forces
Contact forces can be resolved into parts or components
Normal contact force (Normal reaction force)—Line of action of the force is perpendicular to the surfaces in contact
Acts upward on runner and downward on ground
Friction—Line of action of friction is parallel to the two surfaces in contact and opposes motion or sliding between the surfaces
Acts forward on runner and backward on ground

17. Friction
Dry friction—Acts between the non-lubricated surfaces of solid objects or rigid bodies and acts parallel to the contact surfaces
Arises due to interactions between the molecules of the surfaces in contact
Static friction—When dry friction acts between two surfaces that are not moving relative to each other
Limiting friction—Maximum static friction that develops just before the two surfaces begin to slide
Dynamic friction (Sliding friction, Kinetic friction)—When dry friction acts between two surfaces that are moving relative to each other

18. Friction and Normal Contact Force
Self Experiment 1.1
How does adding books to the pile cause the static friction force to increase?
Friction force is proportional to the normal contact force

19. Friction and Normal Contact Force
Self Experiment 1.2
Is friction force only a horizontal force?
Is the normal contact force always vertical and related to the weight of the object that friction acts on?

20. Friction and Surface Area
Self Experiment 1.3
Does increasing or decreasing area in contact affect friction force?

21. Friction and Surface Area
Force pushing the surfaces together remains the same, the force is spread over a greater area
Pressure between the surfaces decreases (pressure equals force/area)
Individual forces pushing each of the molecules together at the contact surfaces decreases
Interactions between the molecules decreases
Friction decreases
An increase in surface area increases the number of molecular interactions, but the decrease in pressure decreases the magnitude of these interactions—The net effect is zero and friction is unchanged

22. Friction and Contacting Materials
What about the nature of the materials in contact?
Rubber soled shoes vs. leather soled shoes

23. Friction and Contacting Materials
Fs = µsR
Fd = µdR
Fs = static friction force
µs = coefficient of static friction
µd = coefficient of dynamic friction
Number that accounts for the different effects that materials have on friction
R = normal contact force

24. Friction in Sport and Human Movement
Static friction is larger than dynamic friction
More difficult to start an object moving than to keep it moving
Locomotion requires frictional forces
Materials used for the soles of athletic shoes have large coefficients of friction
Wax skis to decrease coefficient of friction
Tape, chalk, and sprays improve grip which increases coefficient of friction between hands and implement

25. Addition of Forces: Force Composition
Several external forces can act on us in sports activities
Gravity, friction, normal contact force
In most sport and exercise situations more than one external force will act on an individual
How do we add these forces to determine their effect on a person?
What is the net force or resultant force?

26. Addition of Forces: Force Composition
Full description of force includes magnitude and direction (force is a vector)
Resultant force—Vector addition of two or more forces
Net force—Vector addition of all the external forces acting on an object

27. Colinear Forces
Same line of action—May act in the same direction or in opposite directions
Visually we can think of forces as arrows:
Length of the shaft = magnitude of force
Orientation of the arrow = line of action
Arrow head = direction of action
When forces act along the same line (colinear), they can be added using regular algebraic addition
Add forces to the right (positive) and forces to the left (negative)
Sample Problem 1.1 (text p. 28)
Add forces upward (positive) and forces downward (negative)

28. Concurrent Forces
Do not act along the same line, but do act through the same point
Gymnast example (text p. 29)
How large is the force of gravity that acts on the gymnast?
Force of gravity acting on an object is the object’s weight (W = mg)
What is the net external force acting on the gymnast?
We can represent the forces graphically

32. Concurrent Forces
Resultant directed “upward and slightly to the left” (not very precise description)
Calculate resultant horizontal and vertical forces separately
Left and downward (negative)
Right and upward (positive)

34. Trigonometric Technique
Right triangle—90° angle formed by horizontal resultant force and vertical resultant force
Pythagorean theorem
A2 + B2 = C2
A and B represent the two sides that make up the right angle and C represents the hypotenuse (the side opposite the right angle)
Solve for C to determine the resultant force

35. Trigonometric Technique
Other relationships between sides of right triangle
If we know the lengths of any two sides, we can determine the length of the other side, and the size of the angle between the sides
Conversely, if we know the length of one side of a right triangle and the measurement of one of the angles other than the right angle, we can determine the lengths of the other sides and the measurement of the other angle

36. Trigonometric Technique
Relationships exist between the lengths of the sides of a right triangle and the angles in a right triangle

37. Trigonometric Technique
These relationships can be expressed as ratios of one side to another for each size of angle that may exist between two sides of a right triangle
Sinθ = opposite side/hypotenuse
Cosθ = adjacent side/hypotenuse
Tanθ = opposite side/adjacent side
“Some Of His Children Are Having Trouble Over Algebra”

38. Trigonometric Technique
Theta (θ) represents the angle
Opposite refers to the length of the side of the triangle opposite the angle theta
Adjacent refers to the length of the side of the triangle adjacent to the angle theta
Hypotenuse refers to the length of the side of the triangle opposite the right angle
Sin refers to sine
Cos refers to cosine
Tan refers to tangent

40. Trigonometric Technique
If the sides of the right triangle are known, then arcsine, arccosine, and arctangent functions are used to compute the angle
θ = arcsin (opposite/hypotenuse)
θ = arccos (adjacent side/hypotenuse)
θ = arctan (opposite side/adjacent side)

41. Trigonometric Technique
Gymnast example
θ = arctan (opposite side/adjacent side)
θ = arctan (50N/10N)
θ = arctan(5)—arctan usually abbreviated tan-1
θ = 78.7°
Angles in a right triangle add up to 180°; one angle is 90°, so the sum of the other two angles is 90°
In the gymnast example, what would the other angle be?

42. Trigonometric Technique
Sample Problem 1.2 (text p. 34)
Use the Pythagorean theorem to compute the size of the resultant force
Use the arctangent function to determine the angle of the resultant force with the horizontal

43. Resolution of Forces
What if the forces acting on the object are not colinear and do not act in a vertical or horizontal direction?
Shot-put (Figure 1.1)
Athlete exerts a 100N force on a 4kg shot at an angle of 60° above the horizontal
What is the net force acting on the shot?
First determine the weight of the shot

45. Resolution of Forces
Weight of the shot (40N) acts vertically (downward)
Force from the athlete (100N) acts horizontally and vertically (pushing upward and forward)
Calculate horizontal and vertical components of 100N resultant force separately

48. Resolution of Forces Trigonometric technique
The 100N force represents the hypotenuse acting upward and to the right 60° above horizontal
The side of the triangle adjacent to the 60° angle represents horizontal component
Cosθ = adjacent side/hypotenuse
(100N)Cos60 = adjacent side
Horizontal force component = 50N

49. Resolution of Forces
Side of the triangle opposite the 60° angle represents the vertical component
Sinθ = opposite side/hypotenuse
(100N)Sin60 = opposite side
Vertical force component = 86.6N
Determine net force acting on the shot by adding all the horizontal forces to get the resultant horizontal force and then adding all the vertical forces to get he resultant vertical force

50. Resolution of Forces Only horizontal force acting on the shot is 50N
Two vertical forces acting on the shot
Weight of the shot acting downward (-40N)
Vertical component of the 100N force acting upward (+86.6N)
Add these forces to get 46.6N acting upward
Use Pythagorean theorem to get the net force acting on the shot (Figure 1.13)
(50N)2 + (46.6N)2 = C2
C = √4672N2
C = 68.4N

51. Resolution of Forces
Determine angle between hypotenuse (68.4N) and adjacent (50N) sides
θ = arctan (opposite side/adjacent side)
θ = arctan (46.6N/50N)
θ = 43°
Sample Problem 1.3 (text p. 38)
Calculate horizontal and vertical components of biceps force separately

52. Static Equilibrium
If an object is at rest then the forces acting on the object are in equilibrium and the object is described as being in a state of static equilibrium
Gymnastics coach may want to know how strong a gymnast must be to hold a certain position such as an iron cross
We need to know what external forces are acting on the gymnast

53. Free-Body Diagrams Example 1
A 50kg woman in skates is standing still on ice
What external forces act on her?
Force of gravity acts vertically (downward) through the woman’s center of gravity—an imaginary point in space through which the force of gravity acts on an object
Reaction force from ice acts vertically (upward)

55. Free-Body Diagrams
Free body diagram is a mechanical representation of an athlete—Only the object in question is drawn along with all of the external forces that act on it
Any horizontal forces?
On ice, friction is very small, and because the woman is standing still, the frictional forces under her skates must be zero

56. Static Analysis
If an object is not moving, it is in static equilibrium; acceleration is zero, and the sum of all external forces acting on the object is zero (ΣF=0)
Reaction force of ice exerted on skater
ΣF = R + W = R + (-500N) = 0
R = reaction force from the ice
W = weight of skater
R = 500N

57. Static Analysis
Reaction force from the ice is equal to the weight of the skater but acting in the opposite direction
Example 2
80kg weightlifter has lifted 100kg barbell over his head and is holding it still; he and the barbell are in static equilibrium
What is the reaction force from the floor that must act on the weightlifter’s feet?

58. Static Analysis Three external forces
Reaction force from the floor acting upward
Weight of the athlete acting downward
Weight of the barbell acting downward on the athlete’s hands

59. Static Analysis Reaction force exerted by the hands on the barbell
ΣF = R + W = R + (-1000N) = 0
R = 1000N
Hands exert 1000N upward force on barbell
Reaction force from floor exerted on athlete
ΣF = R + W = R + (-1800N) = 0
R = 1800

61. Static Analysis Example 3
A 20kg child is sitting on a swing, and the child’s parent is exerting a 40N horizontal force on the child to the right and 10N upward force; the child is not moving and is in a state of static equilibrium
What is the resultant force exerted by the swing on the child?
Use one equation for the horizontal forces (ΣFx = 0) and one for the vertical forces (ΣFy = 0)

63. Static Analysis ΣFx = Rx + 40N = 0 Rx = -40N (left)
ΣFy = RY + 10N + (– 200N) (weight of child) = 0
RY = +190N (upward)
Determine the magnitude of the resultant using the Pythagorean theorem
(40N)2 + (190N)2 = C2
C = √37,700N2
C = 194N

64. Static Analysis
Must describe the angle the force makes with horizontal
θ = arctan (opposite side/adjacent side)
θ = arctan (190N/40N) = 78°
Resultant force exerted by the swing on the child is a force of 194N acting backward and upward on the child at an angle of 78° from the horizontal

65. Summary Forces are pushes or pulls
Vector quantities described by size and direction of action
Use arrows to represent forces graphically
Internal forces hold things together and cannot cause changes in motion without external forces
Most common external forces are gravity and contact forces
Friction and normal reaction force are two components of a contact force

66. Summary Forces are added using vector addition
Accomplished using graphical techniques or algebra if the forces are resolved into horizontal and vertical components
If the net external force acting on an object is zero, the object is standing still or moving in a straight line at a constant velocity and is in static equilibrium