20 N 10 N 1m 2m CG The weights are balanced, creating equal torques on either side of the fulcrum. Center of Gravity (COG) zthe point around which a body’s.

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

Center of Gravity (COG)

COG ~ COMGeometric center?Fixed?Located outside of

the body?Moves in the

direction of added mass

Center of Gravity (COG)

COG ~ COMGeometric center?Fixed?Located outside of

the body?Moves in the

direction of added mass

Applications

Muscle Function – Rotational EffectsSegmental AlignmentsWhole Body Stability

Basic Kinetic Concepts

Course Content

I. Introduction to the CourseII. Biomechanical Concepts Related

to Human MovementIII. Anatomical Concepts Related to

Human MovementIV. Applications in Human Movement

Biomechanical Concepts

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

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.

Basic Kinetic Concepts

1. Force2. Torque3. Free Body Diagrams

Basic Kinetic Concepts

1. Force2. Torque3. Free Body Diagrams

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

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.

Common Forces

WeightReaction forces

GRF Friction JRF

Muscle forceElastic forceIntraabdominal pressureInertial forceFluid force

Basic Kinetic Concepts

1. Force2. Torque3. Free Body Diagrams

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)

For a torque to cause angular acceleration:

Must overcome opposing torques (net torque)

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

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

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

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

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.

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

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

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

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

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

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

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

Applications of Torque in Human Movement

Muscle FunctionMovement Analysis

What kind of torque does the biceps brachii create?

What kind of torque does gravity create?

Which way will the arm rotate?

NcmT

NcmNcmT

cmNcmNT

dFdFT

TTT

net

net

net

ggbbbbnet

gbbnet

50

)250(200

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

))(())((

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

What role does the triceps brachii play?

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

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

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

Can this muscle cause frontal plane rotation of the scapula?

Will this muscle cause downward rotation?

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

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

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?

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?

Applications of Torque in Human Movement

Muscle FunctionMovement Analysis

What happens to the resistive force throughout the ROM?

What about the resistive torque?

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

W

d

d

W

W

d

d

W

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

Basic Kinetic Concepts

1. Force2. Torque3. Free Body Diagrams

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.

For the next lecture unit:

Lecture Topic #3 Subtopic A – The Skeletal System