Chapter 1

Forces Maintaining Equilibrium or Changing Motion

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

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

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

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

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

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

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)

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)

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?

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

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

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

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

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?

Friction and Surface Area

• Self Experiment 1.3• Does increasing or

decreasing area in contact affect friction force?

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

Friction and Contacting Materials

• Friction and Contacting Materials

• What about the nature of the materials in contact?– Rubber soled shoes vs. leather soled shoes

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

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

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?

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

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)

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

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)

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

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

Trigonometric Technique

• Relationships exist between the lengths of the sides of a right triangle and the angles in a right triangle

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”

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

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)

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?

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

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

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

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

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

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

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

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

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)

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

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

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?

Static Analysis

• Three external forces 1. Reaction force from the floor acting upward

2. Weight of the athlete acting downward

3. Weight of the barbell acting downward on the athlete’s hands

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

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)

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

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

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

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