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Center of Gravity (COG)

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10 N. 20 N. CG. 2m. 1m. The weights are balanced, creating equal torques on either side of the fulcrum. Center of Gravity (COG). the point around which a body’s weight is equally distributed in all directions. Center of Gravity (COG). COG ~ COM Geometric center? Fixed? - PowerPoint PPT Presentation
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20 N 10 N 1m 2m CG The weights are balanced, creating equal torques on either side of the fulcrum. Center of Gravity (COG) the point around which a body’s weight is equally distributed in all directions
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Page 1: Center of Gravity (COG)

20 N 10 N

1m 2m

CG

The weights are balanced, creating equal torques on either side of the fulcrum.

Center of Gravity (COG)

the point around which a body’s weight is equally distributed in all directions

Page 2: Center of Gravity (COG)

Center of Gravity (COG)

COG ~ COMGeometric center?Fixed?Located outside of

the body?Moves in the

direction of added mass

Page 3: Center of Gravity (COG)

Center of Gravity (COG)

COG ~ COMGeometric center?Fixed?Located outside of

the body?Moves in the

direction of added mass

Page 4: Center of Gravity (COG)

Applications

Muscle Function – Rotational EffectsSegmental AlignmentsWhole Body Stability

Page 5: Center of Gravity (COG)

Basic Kinetic Concepts

Page 6: Center of Gravity (COG)

Course Content

I. Introduction to the CourseII. Biomechanical Concepts Related

to Human MovementIII. Anatomical Concepts Related to

Human MovementIV. Applications in Human Movement

Page 7: Center of Gravity (COG)

Biomechanical Concepts

A. Basic Kinematic ConceptsB. Vector AlgebraC. Basic Kinetic Concepts

Page 8: Center of Gravity (COG)

What is kinetics?

The study of forces tending to cause, causing, or resulting from motion.

Can identify and alter forces to change motion for desirable results.

Page 9: Center of Gravity (COG)

Basic Kinetic Concepts

1. Force2. Torque3. Free Body Diagrams

Page 10: Center of Gravity (COG)

Basic Kinetic Concepts

1. Force2. Torque3. Free Body Diagrams

Page 11: Center of Gravity (COG)

Force

A push or pullVector quantity with 4 characteristics

Magnitude Direction Point of application Line of force

The interaction of an object with its surroundings

Page 12: Center of Gravity (COG)

For a force to cause acceleration:

Must overcome opposing forces (net force)

Must overcome inertia (mass)

Even if acceleration does not occur, deformation of the object will occur.

Page 13: Center of Gravity (COG)

Common Forces

WeightReaction forces

GRF Friction JRF

Muscle forceElastic forceIntraabdominal pressureInertial forceFluid force

Page 14: Center of Gravity (COG)

Basic Kinetic Concepts

1. Force2. Torque3. Free Body Diagrams

Page 15: Center of Gravity (COG)

Torque

AKA moment of force, or momentThe ability of a force to create rotation

Fdmoment arm - perpendicular

distance from the line of force to the axis of rotation

Must have a moment arm (be an eccentric force)

Page 16: Center of Gravity (COG)

For a torque to cause angular acceleration:

Must overcome opposing torques (net torque)

Must overcome moment of inertia (mass, length of the rotating body)

Page 17: Center of Gravity (COG)

Which way will the door turn?

dC = .05 m

FB = 100 N

FC = 200 N

dB = 1 m

TC = FC*dC TB = FB*dB

TC = 200 N * .05 m TB = 100 N * 1 mTC = 10 Nm TB = 100 Nm

TN = 100 Nm + (-10 Nm)TN = 90 Nm

Page 18: Center of Gravity (COG)

dC = .05 m

FB = 100 N

FC = 200 N

dB = 1 m

What must be done for the class to win?

TC or decrease TB sufficiently.

How?

TC FC

dC

Page 19: Center of Gravity (COG)

TC = FC*dC TB = FB*dB

TC = 300 N * .05 m TB = 100 N * 1 mTC = 15 Nm TB = 100 Nm

TN = 100 Nm + (-15 Nm)TN = 85 Nm

Increase the magnitude of FC.

FC = 300 N

Page 20: Center of Gravity (COG)

TC = FC*dC TB = FB*dB

TC = 2000 N * .05 m TB = 100 N * 1 mTC = 100 Nm TB = 100 Nm

TN = 100 Nm + (-100 Nm)TN = 0 Nm

It would take 2000 N of force to get 100 Nm of torque.

Page 21: Center of Gravity (COG)

Increase the magnitude of dC by changing the point of force application.

TC = FC*dC TB = FB*dB

TC = 200 N * .15 m TB = 100 N * 1 mTC = 30 Nm TB = 100 Nm

TN = 100 Nm + (-30 Nm)TN = 70 Nm

FB = 100 N

FC = 200 N

dC = .15 m

Page 22: Center of Gravity (COG)

It would take a moment arm of .5 m to get 100 Nm of torque. We can not apply the force far enough away to get a moment arm this large. We run out of door!

TC = FC*dC TB = FB*dB

TC = 200 N * .5 m TB = 100 N * 1 mTC = 100 Nm TB = 100 Nm

TN = 100 Nm + (-100 Nm)TN = 0 Nm

Page 23: Center of Gravity (COG)

Increase the magnitude of dC by changing the direction of force application.

TC = FC*dC TB = FB*dB

TC = 200 N * .20 m TB = 100 N * 1 mTC = 40 Nm TB = 100 Nm

TN = 100 Nm + (-40 Nm)TN = 60 Nm

Page 24: Center of Gravity (COG)

We would not be able to change the angle enough to get 100 Nm of torque.

Page 25: Center of Gravity (COG)

It will take a combination of all three, or else TB would have to be reduced in a similar (but opposite) manner.

Page 26: Center of Gravity (COG)

How would the door rotate if a force were applied to the door in this manner?

Page 27: Center of Gravity (COG)

Summary

Net torque determines rotationCan change net torque by changing one or

more individual torquesCan change individual torques by

changing Magnitude of force Direction of force (moment arm) Point of application of force (moment arm)

Often takes a combination of all of these

Page 28: Center of Gravity (COG)

Applications of Torque in Human Movement

Muscle FunctionMovement Analysis

Page 29: Center of Gravity (COG)

What kind of torque does the biceps brachii create?

What kind of torque does gravity create?

Page 30: Center of Gravity (COG)

Which way will the arm rotate?

NcmT

NcmNcmT

cmNcmNT

dFdFT

TTT

net

net

net

ggbbbbnet

gbbnet

50

)250(200

)10)(25()2)(100(

))(())((

Page 31: Center of Gravity (COG)

What conditions have to be true for the arm to flex? For the arm to extend?

What role does the triceps brachii play?

Page 32: Center of Gravity (COG)

What is the torque output of this muscle in the frontal plane?

Page 33: Center of Gravity (COG)

What is the torque output of this muscle in the frontal plane?

Page 34: Center of Gravity (COG)

What are the angular movements of the scapula in the frontal plane called?

Can this muscle cause frontal plane rotation of the scapula?

Page 35: Center of Gravity (COG)

Will this muscle cause downward rotation?

Yes, if it has a moment arm for that axis of rotation!

Page 36: Center of Gravity (COG)

Yes, if its torque is larger than the opposing torques.

Page 37: Center of Gravity (COG)

What happens to the length of the moment arm of the muscle throughout the ROM?

What about muscle torque? Does muscle force stay constant through the ROM?

Page 38: Center of Gravity (COG)

What happens to the length of the moment arm of the muscle throughout the ROM?

What about muscle torque? Does muscle force stay constant through the ROM?

Page 39: Center of Gravity (COG)

Applications of Torque in Human Movement

Muscle FunctionMovement Analysis

Page 40: Center of Gravity (COG)

What happens to the resistive force throughout the ROM?

What about the resistive torque?

Page 41: Center of Gravity (COG)

What happens to the resistive force and resistive torque as she goes through the ROM?

W

d

d

W

Page 42: Center of Gravity (COG)

W

d

d

W

What would happen to resistive torque if she put her hands behind her head?

Page 43: Center of Gravity (COG)
Page 44: Center of Gravity (COG)
Page 45: Center of Gravity (COG)

Basic Kinetic Concepts

1. Force2. Torque3. Free Body Diagrams

Page 46: Center of Gravity (COG)
Page 47: Center of Gravity (COG)
Page 48: Center of Gravity (COG)
Page 49: Center of Gravity (COG)

Summary

Net force determines magnitude and direction of acceleration for linear motion.

Net torque determines magnitude and direction of acceleration for angular motion.

FBDs are a necessary first step in analyzing human motion.

Page 50: Center of Gravity (COG)

For the next lecture unit:

Lecture Topic #3 Subtopic A – The Skeletal System


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