CH 13,On BioMechanics in Bio Medical Engineering

7/28/2019 CH 13,On BioMechanics in Bio Medical Engineering

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[BIO-MECHANICS IN BIO-MEDICAL ENGINEERING, CH-

13 , EQUILIBRIUM AND

HUMAN MOVEMENT : ][TEXT BOOK: BASIC-BIO-MECHANICS BY SUSAN-J HALL ,

3rd EDITION ];

6/20/2013

BIO-MEDICAL ENGINEERING GUIDE.INC

MUHAMMAD-SIKANDAR-KHAN-LODHI


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[BIO-MECHANICS IN BIO-MEDICAL ENGINEERING, CH-13 ,

EQUILIBRIUM AND HUMAN MOVEMENT : ]

I [ W W W . M E D I C A L - I M A G E - P R O C E S S N G . B L O G S P O T . C A B Y

M O H A M M A D - S I K A N D A R - K H A N - L O D H I ]

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Table of ContentsEQUILIBRIUM AND HUMAN MOVEMENT: .................................................................................................... 3

1. TORQUE :- ......................................................................................................................................... 3

DIAGRAM: [ fig # 13.1 ( a )] [ Moment-Arm ] ........................................................................................ 4

Fig # 13.2 [ b and c ] .............................................................................................................................. 5

DIAGRAM [ fig # 13.3 ] :- ....................................................................................................................... 6

DIAGRAM [ fig # 13.4 ] [ stage-a ] : ....................................................................................................... 7

DIAGRAM [ fig # 13.5 ] [ stage-b and c ] : .............................................................................................. 8

COUPLING-FORCE-NUMERICAL [ SAMPLE-PROBLEM # 1 ]:- ................................................................... 9

DIAGRAM [ fig # 13.6 ] :- .................................................................................................................... 10

LEVERS :- ..................................................................................................................................................... 12

HUMAN-SYSTEM ACT AS A LEVERS SYSTEM:- ........................................................................................ 12

DIAGRAM [ fig # 13.7 ] First Class of Lever :........................................................................................ 14

Example : [ fig # 13.8 ] ......................................................................................................................... 15

DIAGRAM [ Fig # 13.9 ] ........................................................................................................................ 16

Fig # 13.10 [ a ]: ................................................................................................................................... 17

Fig # 13.10 [ b ] .................................................................................................................................... 18

3rd CLASS OF LEVER:- .......................................................................................................................... 18

Fig # 13.11 : ......................................................................................................................................... 19

Fig # 13.13 : ......................................................................................................................................... 21

MECHANICAL ADVANTAGE:- ................................................................................................................... 21

[ Fig # 13.15 ] ..................................................................................................................................... 24

EQUATION OF STATIC EQUILIBRIUM :- ....................................................................................................... 25

Diagram: [ Fig # 13.16 ] [ Bicep work out ] .......................................................................................... 26

[ Fig # 13.17 (a ) ]. ................................................................................................................................ 27

FIG # 13.17 (b) ..................................................................................................................................... 28

SYSTAMATIC DIAGRAM: [ Fig # 13.19 ( a) ] ........................................................................................ 31

Fig # 13.20 (b) ...................................................................................................................................... 32

Diagram : [ Fig # 13.21 ] ...................................................................................................................... 34

Fig # 13.22 ........................................................................................................................................... 35

Fig # 13.2 ............................................................................................................................................. 36


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[ Fig # 13.26 ] ....................................................................................................................................... 40

DYNAMIC-EQULIBRIUM :- ........................................................................................................................... 43

Fig # 13.29 ........................................................................................................................................... 44

Fig # 13.28 ........................................................................................................................................... 45

EQUILIBRIUM AND HUMAN MOVEMENT:

EQUILIBRIUM:

1.TORQUE :-a.TORQUE [ DEFINITION # a ]:-

The torque [ ] is the product of force [ F ] which acton the body and the moment-arm .

MOMENT-ARM [d ] : Its the perpendicular distance from the force [or

forces ] line of action to the axis of Rotation is called Moment-Arm .

FORMULA:

[ ];

Where :

F = Applied Force .

d = Moment-Arm.


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DIAGRAM: [ fig # 13.1 ( a )] [ Moment-Arm ]

TORQUE : The Rotary effect created by the applied force is known as

torque .


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Fig # 13.2 [ b and c ]

[ ];[ ];

b.TORQUE [ DEFINITION # b ] :-The length of moment-arm and the magnitude of applied force cause

to effect the amount of torque , if we increasing or decreasing thevalue of moment-arm and magnitude of applied force so it cause to

change the value of torque .

c. TORQUE [ DEFINITION # c ]:-


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In this case we applied force on the point of axis of rotation , so then

the moment-arm become zero [ d = 0 ] so, then there is no torque (

Turning effect ) will be produced on that rotator object .

That is :

DIAGRAM [ fig # 13.3 ] :-

Mathematically :

[ ]; {:. [ moment arm = d = 0 ] }[ ];---------finished-here------


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EXAMPLE :

Q) There is a task we have if a person take bicep exercise so, its torque

will be change in each condition , so what we do to maintain some

torque in each complete cycle event of bicep curl?

SOLUTION:

DIAGRAM [ fig # 13.4 ] [ stage-a ] :

[ ];


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DIAGRAM [ fig # 13.5 ] [ stage-b and c ] :

In the complete bicep curl the torque will be change because themoment-arm [ moment arm : the perpendicular distance from

line of action of force to the axis of rotation [ d ]], .

In each position , if moment-arm will be change so, then thetorque will also be changed so if we want to produce the same

torque in each above three stage [a,b,c ] so, we want to prevent

more force F on the stage [ b,c ] where , [ ] AScompared to force [ F1 ] in stage-a because , we do this because in

stage b and c the moment-arm [ d ] will be decreased.

[ ];


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---------------------FINISHED-HERE---------------

COUPLE-FORCE:-

coupling force are those forces in which the two forces which areplaced on both side of the center-of-rotation, and one of these

forces act in the clock-wise direction while the other force act

on the anti [or counter ]-clock wise direction, and it produced the

same effect is known as Coupling-Forces.

---------------------

COUPLING-FORCE-NUMERICAL [ SAMPLE-PROBLEM # 1 ]:-

Q ) The two children sit on a opposite side of a play-ground see-saw. If

joey weight = 200N, is 1.5m from the see-saws axis-of-rotation, and if

Susie weight = 190N, is 1.6m away from the axis-of-rotation so,

which end of the see-saw will drop down towards ground surface?


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DIAGRAM [ fig # 13.6 ] :-

SOLUTION:

{:. Clock wise direction = negative direction }

{:. Counter [ anti ]-clock wise direction = Positive direction }

DATA :

FOR Joey :-

[ ];[ ];


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FOR Susie :-

[

];

[ ];

DISCRIPTION:

As we see in the above diagram that there are coupling forces that is [

Fj = Wj = 200N ] and [ Fs = Ws= 190 N ] act on the Axis of rotation but

in opposite in direction, so, the [ Fs = Ws= 190 N ] act in the counter

[anti ] clock wise [Positive ] direction , also the [ Fj = Wj = 200N ] act in

the clock-wise negative direction , we just sum-up the both torques

produced by the both bodies to find the resulting torque .

1.TORQUE GIVE BY JOEY :-Its a clock-wise direction so, torque is negative

[ ];2.TORQUE GIVE BY THE SUSIE :-

Its a counter-clock wise direction so the torque is positive .

[ ];Now we compare both torque with , so if the magnitude of anytorque is greater than the other one so then as the result of it the seesaw move on that directions , now by adding the eq-i with eq [ ii ] , we

get , the resulting torque we sum-up both torque then we get,

[ ];


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[ ];CONCLUSION:

The torque give by the Susie is greater than the torque give bythe joey , so, as the result of it the torque which give by Susie causeto rotate the see-saw on the counter-clock wise [ positive ] direction .

--------------finished-here-----

LEVERS :-

The liver is a Rigid-bar , which can rotated around its axis under the

action of some external forces.

HUMAN-SYSTEM ACT AS A LEVERS SYSTEM:-

In the human body there are the some biological liver are presentspecially occurs in the limbs.

In the Biological system the both limbs act as a Livers .so as we consider the upper-limbs, in the Upper-Limbs the

Humerious [bone of arm ] and the radius + ulna [ both bones

of fore-arm ] act as a rigid bar and the joint is act as the axis-of-

rotation and the bi-cep and tri-cep act as a applied force on

the rigid-bars of the Levers.


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there are three types of Levers which are given below.

That is :

1.1st Class of Levers:2.2ndClass of levers.3. 3rd Class of Levers.

1.1st CLASS OF LEVERS :-

In the 1

st

class of levers , the applied force [ F ] and the resistanceforce [ R ] are located on opposite site of levers .

This applied force [ F ] and Resistance force [ R ] are both in theopposite in direction one force is in the clockwise [ negative ]

direction and the other 2nd

force is in the counter-clock wise in the

positive direction.

There is an another possibility is that the both forces Appliedforce [ F ] and the resistance force [ R ] in equal in distance withrespect to the Axis of rotation, with each other .

Or , its also be a possibility that the one force is approximatelynear [ or far away ] with respect to the axis as compared to the 2

nd

force.


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DIAGRAM [ fig # 13.7 ] First Class of Lever :


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Example : [ fig # 13.8 ]

2.2nd CLASS OF LEVERS :-The applied force [ F ] and resistance force [ R ] on same side of liver in

which the resistance [ R ] is lie near to the Axis-of-rotation .


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DIAGRAM [ Fig # 13.9 ]


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Fig # 13.10 [ a ]:


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Fig # 13.10 [ b ]

3rd CLASS OF LEVER:-

In the 3rd

class of lever the applied force [ F ] and resistance [ R ] are lie

on same side of lever where force [F ] lie too near to the Axis-of-

Rotation.

Example:


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Fig # 13.11 :

Diagram: [ fig # 13.12 ]


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-------------finished here ----------

Example :


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Fig # 13.13 :

When ever the moment arm of applied force is greater then the

moment arm of resistance [ R ] so, the magnitude of applied force [ F ]

is smallest then the magnitude of resistance [ R ].

--------------finished here ------------

MECHANICAL ADVANTAGE:-

The mechanical advantage is the ratio of moment arm of applied force

[F ] over the moment arm of resistance [ R ].

FORMULA :

[ -> Mechanical-Advantage = moment arm [ F ] / moment arm [ R ] ];


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[ ];[

];

{:. its mean much less then, or vice verca };Fig # 13.14 [ MECHANICAL ADVANTAGE ]:

Example : [ Mechanical-Advantage

1 ]:

When the moment-Arm of force [ F ] is greater then themoment-Arm of the resistance [ R ] , then the mechanical-

advantage is equal to greater then one.


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Because , this value of mechanical advantage shows us that whichis a greater then one [ Mechanical advantage 1 ] .

We cause to applied less magnitude of force [ F ] then ascompared to the force of the resistance , because this greater

moment arm of the force [ F ] give the greater amount of torque

which cause to pull the heavy resistance force [ R ] so, its the

mechanical advantage for the Applied force [ F ] .

----------------finished ---

EXAMPLE-2 : [ MECHANICAL ADVANTAGE ] :

[ ];[ ];

Condition => [ Mechanical Advantage 1 ];


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[ Fig # 13.15 ]

In the above condition when the moment arm [ ] is smallerthen the moment arm [ ] , then , if we want to move the leverso, we should be provided to more forces [ greater force ] , Then

the resistance so the Mechanical advantage is much less than [

] 1 .This value of mechanical advantage much less than one indicate

that this arrangement is less effective in the sense we provided

greater external applied force to move the lever , [ as we compare

to the previous example].

---------------------finished here -------------


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EQUATION OF STATIC EQUILIBRIUM :-

When ever the body is in the state of motion less or in rest state [ or

its in the state of static ] , so then this body in the state of Static-Equilibrium , also , there are three condition to set the body in the

state of Static-Equilibrium , that is .

1.The sum of all Vertical-forces act on that body is equal to zero .2.The sum of all horizontal forces act on that body is equal to

zero .

3.The sum of all torque acting on the body is equal to zero.[ ];

[ ];[ ];Key point => if any one of above these three condition is not equal to

zero , so then this body is in the state of Motion .

------------finished here------------

SAMPLE PROBLEM-2 [ STATIC-EQULIBRIUM ] :-

Q ) how much force must be produced by the bicep brachii , attaching

at 90 degree to the radius at 3cm from the center of rotation at the

elbow joint to support a weight of 70N held in the hand at a distance of

30cm from the elbow joint ?

Note : we neglect the weight of the fore arm and hand , and neglected

any action of other muscle ( Triceps).

Solution:


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Diagram: [ Fig # 13.16 ] [ Bicep work out ]

DATA :

Its the 2nd class of lever, in which the bicep muscle act as aresistance against the external applied force [ F ] .

Where , this condition is the static condition , so all the sum oftorque forces acting on that body is equal to zero .

[ ];[ ];


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R = ?

F = external applied force = 70 N ;

[ ];SYSTAMATIC-DIAGRAM:-

[ Fig # 13.17 (a ) ].


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FIG # 13.17 (b)

Req :

R=?

SOLUTION:

As we know that given in the question that the body segment in the

state of Static-Equilibrium, so all the sum of torque forces whichacting on that body segment is equal to zero.

[ ];Formula:


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[ ];FOR R :-

[ ];[ ];[ ];[ ];..

R = 700 N . Answer .

--------------finished-here--------

SampleProblem 3 [ Static-Equilibrium ]

Q ) Two individuals apply force to opposite side of a frictionless

swinging door .if A applied a 30N force at a 40 degree 45 cm from

the doors hinge and B applied force at a 90 degree angle 38cm fromthe doors hinge, what amount of force is applied by B if the door

remains in a static position?

Diagram : [ Fig # 13.18 ]


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SOLUTION:


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SYSTAMATIC DIAGRAM: [ Fig # 13.19 ( a) ]

[ ];Required :

[ ];


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Fig # 13.20 (b)

[ ];So,

Key point: As we know that the body segment is in the static-position

, so the sum of all the Equivalent torque [ ] forces is equal to zero,[ ];Formula:

[ ];Where :


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[ ];[

];

--------------

Plus sign shows that the counter-Clock wise [ Positive ] direction .

And

Minus sign shows that the clock wise direction.

------------

Now placing the above value in eq-A, then we get.

[ ];[ ];[ ];[ ];[ ];..

[ ]; Answer . [ its the Force applied by the B bywhich the body segment stay in the static-position.

CONCLUSION:

We want to provide the FB = 35.52 N to the perpendicular to the door

so, then the door will stay in the static position.

---------------finished-here------------


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Sample-problem [ STATIC-EQULIBRIUM]:

Q) The quadriceps tendon attaches to tibia at a 30 degree angle 4cm

from the joint center at the knee , when an 80N weight is attached to

the ankle 28cm from the knee joint , how much force is required of the

quadriceps to maintain the leg in a horizontal position ( or in static-

position) ?

a. what is the magnitude and direction of the reaction force , exerted

by the femur on tibia?

Note: Neglected the weight of the leg and the action of other muscles.

Diagram : [ Fig # 13.21 ]


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SOLUTION:

Fig # 13.22

FORMULA:

[ ];


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Fig # 13.2

[ ];[ ];

[ ];So,

SOLUTION:


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SCHEMATIC-DIAGRAM: [ Fig # 13.24 ]

REQUIRED :-

PART (a) :

[ ];PART (b) :

Magnitude and direction of force by the femer on the tibia => R = ? , QR

=? .

Part (a) :-


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For FQ:

FREE-BODY DIAGRAM OF FQ :- [ Fig # 13.25 ]

So, this body segment is in the state of static-equilibrium, so, sum of all

forces torque are equal to zero.

[ ];Part-a :

For FQ:-

[ ];


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[ ];{:.

}

[ ];[ ];.

[ = 1120 ] ; its the required torque force for the static-condition.

PART b :

FOR MAGNITUDE AND DIRECTION OF FORCE [ R ] BY THE femer ON

tibia QR :-

FOR R :-

FREE BODY DIAGRAM OF FEMER EXERTING FORCE ON THE TIBIA:


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[ Fig # 13.26 ]

FOR Rx :-

ACCORDING TO THE STATIC-EQUILIBRIUM :

[ ]; sum of all forces in the x-component is zero.[ ];[ ];[ ];[ ];


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FOR Ry :-

According to static equilibrium

[ ]; sum of all forces in the y-component is zero.[ ];

[ ];[ ];Pythagoreous theorem:

[ ];

[ ];

[

];

[ ];[ ]; Answer-1 , its the magnitude of the resultantforce.

FOR-DIRECTION-OF-R :-

Fig # 13.26


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[ ];[ ];

[ ];So,

[ ];[ ]; its the direction of R , Answer-2.---------finished-here---


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DYNAMIC-EQULIBRIUM :-

when any body in the state of motion is known as Dynamic-

equilibrium

so, in the dynamic equilibrium there are all types of the forces act in the

body that is ,

[ sum of all forces acting in a body in x-axis ];[ sum of all forces acting in a body in y-axis ];

[ ];----finished-here--------

CENTER-OF-GRAVITY:-

[ OR CENTER-OF-MASS = CENTER-OF-GRAVITY ]:

CENTER-OF-MASS :-

There is a point in every body where the body mass are equally

distributed in all direction, this point is known as center-of-mass or

Mass-centroid.

CENTER-OF-GRAVITY:-

If we analysis this body in the gravitation forces , so, the center-of-

mass of the body is also known as center of gravity,

Key-point=> the center-of-gravity is that point on the body where sum

of all the torque produced by weight of the body segment is equal to

zero.


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Fig # 13.29

[ ];[ ];[ ];[ ]; net torque is zero.


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Fig # 13.28

-------------finished here---------

----------CH-13 ON BIO-MECHANICE IN BIO-MEDICAL ENGINEERING

FINISHED-HERE----------------

XMUHAMMAD-SIKANDER-KHAN-LODHI

OWNER-OF-MY-PERSONAL NOTES Thursday, June 20, 2013


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CH 13,On BioMechanics in Bio Medical Engineering

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