© 2008 McGraw-Hill Higher Education. All Rights Reserved. Chapter 12: The Conditions of Linear Motion KINESIOLOGY Scientific Basis of Human Motion, 11

© 2008 McGraw-Hill Higher Education. All Rights Reserved.

Chapter 12:The Conditions of

Linear Motion

KINESIOLOGY

Scientific Basis of Human Motion, 11th edition

Hamilton, Weimar & LuttgensPresentation Created by

TK Koesterer, Ph.D., ATC

Humboldt State University

Revised by Hamilton & Weimar

KINESIOLOGY

Scientific Basis of Human Motion, 11th edition

Hamilton, Weimar & LuttgensPresentation Created by

TK Koesterer, Ph.D., ATC

Humboldt State University

Revised by Hamilton & Weimar


Objectives

1. Name, define, and use the terms of linear motion.

2. Define magnitude, direction, and point of application of force and use terms properly.

3. Explain changes magnitude, direction, and point of application of force on the motion state of a body.

4. Define and give examples of linear forces, concurrent forces, and parallel forces.

5. Determine magnitude, direction, and point of application of muscles forces.

6. State Newton’s laws as they apply to linear motion.


Objectives

7. Explain cause and effect relationship between forces of linear motion and objects experiencing the motion.

8. Name & define basic external forces that modify motion.

9. Draw and analyze a 2D free-body diagram.

10. Explain the work-energy relationship applied to a body experiencing linear motion.

11. Define and use properly the terms work, power, kinetic energy, and potential energy.

12. Perform a mechanical analysis of a motor skill.


THE NATURE OF FORCE

• Force is that which pushes or pulls through direct mechanical contact or through the force of gravity to alter the motion of an object.

• Internal forces are muscle forces that act on various structures of the body.

• External forces are those outside the body:– Weight, gravity, air or water resistance,

friction, or forces of other objects acting on the body.


Aspects of Force

• Force is a vector quantity:– Magnitude and direction– Also has a point of application

• All three characteristics must be identified. – For a weight lifter to lift a 250 N barbell:

• Lifter must apply a force greater than 250 N, in an upward direction, through the center of gravity of the barbell.


Magnitude

• Amount of force being applied.– Force exerted by the barbell had a

magnitude of 250 N.– This force was the result of gravity acting

on the mass of the barbell.• In this case, the force is referred to as weight.• Weight is mass times acceleration due to

gravity:w = mg


Magnitude of Muscular Force

• In direct proportion to the number & size of fibers contracting in a muscle.

• Muscles normally act in groups whose force or strength is measured collectively.

• Maximum muscular strength is measured by a dynamometer.

• Measures force applied by a group of muscle through an anatomical lever.


Point of Application

• Point at which force is applied to an object.• Where gravity is concerned this point is

always through the center of gravity. • For muscular force, this point is assumed to

be the muscle’s attachment to a bony lever.• Technically, it is the point of intersection of

the line of force and the mechanical axis of the bone.


Direction

• Direction of a force is along its action line.– Gravity is a downward-directed vector

starting at the center of gravity of the object.

– Direction of a muscular force vector is the direction of line of pull of the muscle.


Direction of Muscular Force Vector

• Muscle angle of pull: the angle between the line of pull and the mechanical axis of the bone.

Fig 12.1


Resolution of Forces

• Magnitude is line A.• Point of application is at

point B.• Direction is represented

by the arrowhead and the angle .

Fig 12.2


Angle of Pull

• Force may be resolved into vertical and a horizontal components.

• Size of each depends on angle of pull.• Since a muscle’s angle of pull changes with

every degree of joint motion, so do the horizontal & vertical components .

• The larger the angle (0° - 90°), the greater the vertical and less the horizontal component.


Angle of Pull

• The vertical component is perpendicular to the lever, called rotary component.

• The horizontal component is parallel to the lever and is the nonrotary component.

• Most resting muscles have an angle of pull < 90°.

Fig 12.1a

Rotary component

Nonrotary component


Rotary vs. Nonrotary Components

Angle of pull < 900

• Nonrotary force is directed toward fulcrum.

• Helps maintain integrity of the joint (stabilizes).

Fig 12.1a

Rotary component

Nonrotary component



Angle of pull > 900

• Dislocating force is directed away fulcrum.

• Does not occur often.• Muscle is at limit of

shortening range and not exerting much force.

Fig 12.1c



Angle of pull = 90°

• Force is all rotary.

Angle of pull = 45°

• Rotary & nonrotary components are equal.

Muscular force functions:• Movement• Stabilization

Fig 12.1b


Anatomical Pulley

• Changes the angle of pull of the muscle providing the force.

• This increase in angle of pull increases the rotary component.– e.g. Patella for the

quadriceps.

Fig 12.4


Resolution of External Forces

• Accomplished in the same manner as muscular forces applied at an oblique angle.

• Only horizontal force will move the table.

• Vertical force serves to increase friction.

Fig 12.7


Composite Effects of Two or More Forces

• Two or more forces can be applied to objects.– A punted ball’s path is the result of force of

the kick, force of gravity, and force of wind.– Muscles work in groups, e.g. the 3

hamstrings.• Composite forces on the body may be

classified according to their direction and application as linear, concurrent, or parallel.


Linear Forces

• For forces applied in the same direction, the resultant is the sum of the forces:

a + b = c

• For forces applied in the opposite directions, the resultant is the sum of the forces:

a + (-b) = c

=+a b c

=+a b c


Concurrent Forces

• Act at the same point of application at different angles.

• Resultant of two or more concurrent forces depends on both the magnitude of each force and the angle of application. Fig 12.8


Parallel Forces

• Forces not in the same action line, but parallel to each other.

• Three parallel forces:– two upward– one downward

Fig 12.9


Parallel Forces

• 10 N weight at 90°.

• Gravity acts at points B & C.

• A is the force of biceps.• Effect of parallel forces

on an object depends on magnitude, direction & application point of each force.

Fig 12.10


NEWTONS’ LAWS OF MOTIONLaw of Inertia

A body continues in its state of rest or of uniform

motion unless an unbalanced force acts on it.– An object at rest remains at rest.– An object in motion remains in same motion– Unless acted upon by an outside force.

• Friction & air resistance effect objects in motion.


Law of Inertia

• A body continues in its state of rest or of uniform motion unless an outside, unbalanced force acts on it.

Fig 12.11

Gravity

Vy

Vx


Law of Acceleration F = ma

The acceleration of an object is directly proportional to the force causing it and inversely proportional to the mass of the object.

What is the force needed to produce a given linear acceleration?

• Since m = w/g, F = (w/g) x a• Force to accelerate a 300 N object 2 m/sec2

• F = (300 N / 9.8m/s2) x 2 m/s2 = 61 N


Impulse Ft = m(vf – vi)

The product of force and the time it is applied.F = ma

• Substitute (vf – vi) / t for a:

F= m (vf – vi) / t• Multiply both sides by

time:

Ft = m (vf – vi)

Fig 12.12


Momentum Ft = mvf - mvi

The product of mass and velocity• 20 N force applied for 5 sec has equal

momentum to a 100 N force falling for 1 sec. Why?

• Any change in momentum is equal to the impulse that produces it.

• Force applied in direction of motion will increase momentum.

• Force applied opposite to direction of motion will decrease momentum.


Law of Reaction

For every action there is an equal and opposite reaction.

Fig 12.13 & 12.14


Conservation of Momentum

In any system where forces act on each other the momentum is constant.

• An equal and opposite momentum change must occur to object producing reactive force.

• Therefore:

m1vf1 – m1vi1 = m2vf2 – m2vi2

Fig 12.15


Summation of ForcesForce generated by muscle may be summated from one

segment to another.

Typical throwing pattern• Force from legs is transferred to the trunk.• Further muscular force increases momentum and is

transferred to upper arm.– Mainly as an increase velocity because mass is

smaller.• Sequential transfer of momentum continues with mass

decreasing and velocity increasing. • Finally momentum is transferred to thrown ball.


FORCES THAT MODIFY MOTIONWeight

• The force of gravity is measured as the weight of the body applied through the center of gravity of the body and directed toward the earth’s axis.

W = mg

Fig 12.16

Weight


Contact Forces:Normal Reaction

• For every action there is an equal and opposite reaction.– The jumper pushes

off the ground and the ground pushes back.

Fig 12.17

Reaction

Action


Contact Forces:Friction

• Friction is the force that opposes efforts to slide or roll one body over another.– In some cases we try to increase friction

for a more effective performance.– In other cases we try to decrease friction

for a more effective performance.• The amount of friction depends on the nature

of the surfaces and the forces pressing them together.


Friction

Friction is proportional to the force pressing two surfaces together.

• Force of friction acts parallel to the surfaces and opposite to the direction of motion. W = weight

T = reactive force of tableP = force needed to moveF = force resisting motion

Fig 12.18


Coefficient of friction,

• The ratio of force needed to overcome the friction, P, to the force holding the surface together, W:

= P / W or

= Fmax/FN– Large coefficient surfaces cling together.– Small coefficient surfaces slide easily.– Coefficient of 0.0 = frictionless surface.


Coefficient of Friction• May be found by:

– Placing one object on a second and tilt the second until first begins to slide.

– The tangent of the angle with horizontal is the coefficient of friction.

Fig 12.19


Elasticity and Rebound

• Objects rebound in a predictable manner.• The nature of rebound is governed by

elasticity, mass, and velocity of rebounding surface, friction between surfaces, and angle of contact.

• Elasticity is the ability to resist distorting influences and to return to the original size and shape.


Elasticity and Rebound

• Stress is the force that acts to distort.

• Strain is the distortion that occurs.

• Stress may take the form of tension, compression, bending, or torsion.

Fig 12.21b


Coefficient of Elasticity

• Is defined as the stress divided by the strain.• Most commonly determined in the

compression of balls by comparing drop height with the bounce height.

• The closer to 1.0 the more perfect the elasticity.

e = bounce height drop height


Coefficient of Elasticity

• Also may be found using the Law of Conservation of Momentum: – Using the change in velocity of the two

objects, assuming masses remain constant:

Where vf2 and vf1 are velocities after impact, and vi1 and vi2 are velocities before impact.

e = (vf1 – vf2) / (vi1 – vi2)


Angle of Rebound• For a perfectly elastic object, the angle of

incidence (striking) is equal to the angle of reflection (rebound).

• As coefficient of elasticity varies variations will occur.

Fig 12.22


Effects of Spin on Bounce

• A ball with topspin will rebound from horizontal surface lower and with more horizontal velocity.

• A ball with backspin will rebound higher and with less horizontal velocity.

• A ball with no spin will develop topspin.• A ball with topspin will gain more topspin.• A ball with backspin may be stopped or reversed.• Spinning balls hitting vertical surfaces will react in the

same manner as with horizontal surfaces, but in relation to the vertical surface.


Fluid Forces

• Water and air are both fluids and as such are subject to many of the same laws and principles.

• The fluid forces of buoyancy, drag, and lift apply in both mediums and have considerable effect on the movements of the human body.


Buoyancy

• Archimedes’ Principle states: a body immersed in a liquid is buoyed up by a force equal to the weight of the liquid displaced.

• This explains why some things float and some things sink.

• Density is a ratio of the weight of an object to its volume.


Specific gravity

• Ratio of the density of an object to the density of water.

• An object the same weight for volume as water has a specific gravity of 1.0.

• An object with specific gravity > 1.0 will sink.• An object with specific gravity < 1.0 will float.


Lift and Drag

Drag is the resistance to forward motion through a fluid.

Result of :• fluid pressure on the

leading edge of the object.

• amount of backward pull produced by turbulence on the trailing edge.

Fig 12.24 b


Lift and Drag

Laminar flow is a smooth, unbroken flow of fluid around an object.

• A smooth surface will have better laminar flow than a rough surface, resulting in less drag.

Fig 12.24 a


Lift and DragLift is the result of changes in fluid pressure as the

result of difference in air flow velocities.

Bernoulli’s Principle states: the pressure in a moving fluid decreases as the speed increases.

Fig 12.24 c

V P

V P Lift

Drag


Ball Spin (Magnus Effect)• Bernoulli’s Principle

applies here also.• A ball will move in the

direction of least air pressure.

• A ball spinning drags a boundary layer of air with it, causing air to move faster & reducing pressure on one side.

Fig 12.25


FREE BODY DIAGRAMS

• In analyzing any technique, one should consider all external forces, by accounting for effect of each one of the body.

• The isolated body is considered a separate mechanical system.

• Easier to identify forces & represent as vectors.

• Can help determine the application and direction of forces acting on the body.


Direction & Point of Application of External Forces

Force Direction of Force Point of Application

Weight (W) Downward Center of Gravity

Normal (R) Perpendicular Point of contact

Friction (F) Along surface Point of Contact

Buoyancy (B) Upward Center of buoyancy

Drag (D) Opposite flow Center of Gravity

Lift (L) Perpendicular to drag Center of Gravity


Free Body Diagram• Magnitude

– arrow length• Direction

– arrow head• Point of application

– arrow tail• Weight (W)• Reactive force (R)• Friction (F)

Fig 12.26


Free Body Diagram

• Weight (W)• Buoyancy (B)• Lift (L)• Drag (D)

State of motion or rest of the body depends on the vector sum of all these forces.

Fig 12.27


Free Body Diagram• Also used to show forces

on a body segment.• Thigh is isolated:

– Weight of thigh (W)

– Muscle force Hip (MH)

– Reactive Forces

• Hip (Hx & Hy)

• Knee (Kx & Ky))

Fig 12.28


WORK, POWER, AND ENERGYWork

• Work is the product of force expended and the distance over which force is applied.

W = Fs– Work (W), Force (F), Distance (s)

• Units are any combination of force & distance:– foot/pounds,– joule = 107 x 1 gram / 1 centimeter


Work• A 20 N suitcase is place on a shelf 2 m above

the ground:– Work done against gravity= 40 Nm

• Same suitcase lifted along a 4 m incline is still 40 Nm of work against gravity.– Horizontal distance not included.

30o

4 m2 m


Positive & Negative Work

• Positive work – force acts in the same direction to that of the objects motion.

• Negative work – force acts in the opposite direction to that of the objects motion.


Mechanical Muscular Work

Example: a rectangular muscle 10 cm x 3 cm, that exerts 240N of force.

• Average muscle fiber shortens 1/2 its length.

W = Fs

W = 240N x 5cm

W = 1200N•cm or 120 Nm


Force per Muscle Cross Section

• If force of the muscle is not known, it is computed form the muscle’s cross section.

Example: Assume same muscle is 1cm thick:

Cross section = width x thickness

3 cm X 1 cm = 3 sq cm

Average force = 360 N per sq cm

F = 360 x 3 = 1080N

W = Fs

W = 1080N x 5cm = 5400 N cm or 540 Nm


Muscular Work

• If the internal structure of the muscle is rectangular, a simple geometric cross-sectional measure can be used.

• For penniform & bipenniform muscle, physiological cross section must be determined.

• “s” represents 1/2 the length of the average fiber.

• Force per square inch depends on whose research the student accepts.


Muscle Work by Physiological Cross Section (PCS)

W = Average force x PCS (sq cm) x .5 length of fibers (cm)

Divide by 100 to convert N-cm to Nm

W (Nm) = 360 x PCS (sq cm) x .5 fiber length (cm)

100


Power

• The rate at which work is done.

P = Fs / t or P = W / t or P = Fv

P = Power t = time

W = work v = velocity = s / t


Energy

• The capacity to do work.

Law of Conservation of Energy:The total amount of energy possessed by a body or an isolated system remains constant.


Potential Energy

• Potential energy: energy based on position.• Potential energy is the product of the weight

of an object and the distance over which it can act:

PE = mgh

m = mass, g = gravity, h = height


Kinetic Energy

• Energy based on motion:

KE = 1/2 mv2

m = mass, v = velocity

• Work done is equal to the kinetic energy acquired, or

Fs = 1/2 mv2


ANALYSIS OF LINEAR MOTION

• First identify the nature of the forces involved in the motion of interest:

– Weight

– Propulsive forces

– Ground Reaction Force

– Friction

– Buoyancy, Drag, & Lift


ANALYSIS OF LINEAR MOTION

• The principles that govern the mechanical aspects of a movement can be summarized by examining some of the basic concepts involved in the kinetics of linear motion:– Inertia– Impulse– Work & Power– Potential & Kinetic Energy


Chapter 12:The Conditions of

Linear Motion

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© 2008 McGraw-Hill Higher Education. All Rights Reserved. Chapter 12: The Conditions of Linear Motion KINESIOLOGY Scientific Basis of Human Motion, 11