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Chapter 51

The Science of Levers__________________________________________________________________

Tommy Boone, PhD, MPH, FASEP, EPC and Larry Birnbaum, PhD, MA, EPC

Three points determine the class of a lever, the axis (pivot, fulcrum), the resistance,and the force applied to the lever. The arrangement of these points determines theclass of lever.

First-class lever the axis (A) is in the middle between the force (F) andresistance (R). The force and resistance are on opposite ends. A first class leverfavors balance when the axis is in the middle. If the axis moves closer to theresistance arm, force is favored (the force arm is longer than the resistance arm).

The longer the force arm, the easier the work is (more force is applied with thesame muscle exertion). Most efficient class of levers. Examples: seesaw, hammerremoving a nail. Force arm may also be referred to as the force point arm;resistance arm as resistance point arm.

F R F R

F Hammer removing a nailR

A

Second-class lever R and F are on the same side of the axis. The F arm islonger than the R arm. Second class levers favor force (power). Examples: push-ups, wheelbarrow, nutcracker.

F AR R

F R A(c/g) A F

Third-class lever- R and F are on the same side of the axis. The R arm is longerthan the F arm. Third class levers favor speed. Examples: shovel, baseball bat,fishing rod, canoe paddle, nearly all muscles of the body (force is the contractingmuscle).

A AR

F F FR F A A

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RR

A A A A AF F F F FR R R R R

A F RBiceps brachii Deltoid Quads Iliopsoas Gluteus maximus Hamstrings

Third class levers depict force, but favor speed. The tennis racket example belowshows that angular velocity is greatest for arc A because there is greaterdisplacement in the same period of time.

A F R Angular velocity = angular displacement time

CB

A

Different class levers provide increases or decreases in three mechanical variables:

1. Force2. Speed of movement3. Range of motion

A decrease in one of these variables is accompanied by an increase in one or both

of the other variables. For example, if the force (F) arm is longer than theresistance (R) arm in a first-class lever, a gain in force results. However, the R armmoves a shorter distance, and thus slower, than the F arm. Hence one variableincreases, two decrease. If the R arm is greater than the F arm, a gain in speedand range of motion results, but force is reduced.

The F arm is always longer than the R arm in second-class levers. Thus, itproduces a gain in force. Because a greater F arm exists, less force is needed toovercome a greater resistance.

The R arm is longer than the F arm in third-class levers. Consequently, there is a

reduction in force, but a gain in speed and range of motion. If a force greater thanthe resistance is applied, the R arm will move rapidly and through a range greaterthan the F arm.

To summarize, the two major purposes that levers serve are to either 1) increaseforce, or 2) increase speed. If the F arm is longer than the R arm, the function ofthe lever is to increase force (second-class). If the F arm is shorter than the R arm,the lever serves to increase speed (third-class). If both arms are of equal length

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(first-class), no advantage is gained by using the lever it does not increase speedor force. Since most of the bony levers have a F arm that is shorter than the R arm(i.e., origin with respect to insertion), the human body is generally considered to bemuch better equipped to make fast movements than forceful ones. The mostcommon human body lever is the third-class one. A couple of exceptions should be

noted.

The triceps brachii is a first-class lever. The force point is the insertion on theolecranon process of the ulna. The resistance point is the c/g of the forearm. Theaxis is the elbow. Since the R arm is greater than the F arm, the triceps brachiiyields a gain in speed and range of the hand with a reduction in force.

Triceps brachii- elbow extension- first-class lever

F A R

Brachioradialis- elbow flexion- second-class lever

A R F

Biceps brachii or brachialis- elbow flexion- third-class lever

A F R

The brachioradialis is a second-class lever. The force point is the insertion on theradial styloid process. The resistance point is the c/g of the forearm. The axis isthe elbow. Since the F arm is longer than the R arm, this muscle produces a gain inforce with a reduction in speed and range.

The biceps brachii and brachialis are third-class levers. The force point for thebiceps brachii is the insertion on radial tuberosity. The resistance point is the c/g ofthe forearm. The axis is the elbow. Such a lever suffers a reduction in force with again in range and speed. Since these muscles are strong enough to overcome the

forearm, they result in increased speed and range of the hand.

Adding external weights to a body segment can cause a shift in the location of itsc/g (i.e., resistance point), which can result in a change in the class of lever. Forexample, without an external weight, the brachioradialis is a second-class lever.With an external weight, it becomes a third class lever.

A F R

The addition of a weight (as when lifting weights) changed thebrachioradialis from a muscle that produced a gain in force toone that results in a gain in speed and range. As such, thebrachioradialis works with or compliments the actions of the

biceps brachii and brachialis muscles.

Positioning the body (either consciously or unknowingly) may result in changing thelever classification that may be either positive or negative. For example, the foot isused as a second-class lever when one rises on the toes (i.e., heel raises). Whenthe plantar flexors contract, the toes become the fulcrum (axis), the body weight is

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the resistance, which lies somewhere between the force point and axis. This leverfavors force.

By leaning forward, the weight is moved closer to the axis resulting ina longer F arm. The shorter the R arm, the less force is needed to liftthe weight. This might be the desired technique if one wants to liftmaximum weight. The coach might instead want to stress thegastrocnemius and soleus (and plantaris) muscles by increasing the Rarm. This would occur when the weight is behind the heels.

F R ALeaningforward

FR ALeaningbackward

F A RExtremeforward

In an extreme situation, the person mightlean forward sufficiently to position theweight (i.e., c/g) in front of the toes, the calfmuscles then function as a first-class lever.Since the F arm is greater than the R arm(in length), the change in class does not

affect the force output.

Some athletes may be endowed with muscle insertions that are farther from theirjoints than in the average person. Only a small difference is necessary to giveconsiderable mechanical advantage. Consider the third-class lever, for example.

Although the R arm will always be longer, positioning the hand close to theresistance makes the resistance easier to handle. Placement of the hand close to

the axis makes the resistance much more difficult to overcome. In this fashion,insertions further down the R arm allow better control/power of the limb, thus,strength; however, a short F arm is necessary if great speed is desired of the limb.

A F R A F R

A F R AF R

Consider the following equation, relative to all three classes of levers:

F x fa = R x ra

F and R refer only to rotarycomponents of the actual

F = product of the applied forcefa = length of the force armR = product of the resistance forcera = length of the resistance armF x fa = force torque

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forces. R x ra = resistance torque

Consider a first-class lever (triceps brachii). Given the following information, what isthe rotary force (or force torque) of contraction?

10 lb force against table top

F AR

R arm = 12 inches (i.e., palm is 12 inches from axis)F arm = 1 inch

F x 1 inch = 10 lbs x 12 inchesF = 120 lbs

Consider a third-class lever (biceps brachii). Given the following information, whatis the rotary force (or force torque) of contraction?

16 lb shot in the hand

A FR

R arm = 14 inches (i.e., from axis to center of shot)F arm = 1 inch

F x 1 inch = 16 lbs x 14 inchesF = 224 lbs

What if the insertion of the biceps brachii is moved 1 inch further from the elbow (Farm = 2 inches)?

F x 2 inches = 16 lbs x 14 inches; F = 112 lbs

Thus, the biceps brachii is more or less efficient depending on its insertion point.How heavy a dumbbell can Tim and Tom hold if they both have 12 inch forearmsand produce 200 lbs of force, but Tims biceps brachii inserts 2 inches from theelbow, whereas Toms insertion is 1.5 inches from the elbow?

Tim:

R = (200 lbs x 2 in) 12 in = 33.3 lbs

Tom:

R = (200 lbs x 1.5 in) 12 in = 25 lbs

Rotational (vertical) component vs. stabilizing (horizontal) component of forceThe force exerted by a muscle to move an object (e.g., weight) can be broken down to arotational or vertical component (RC) and a stabilizing or horizontal component (SC). The

rotational component is perpendicular to the bone serving as a lever and tends to rotate thebone while the stabilizing component is parallel to the bone and compresses the joint

surfaces when the insertion angle () is 90.

When the angle is >90, the stabilizing component may be referred to as the dislocating

component.

VV = VO x sin HV = VO x cos

RC = FO x sin SC = FO x cos

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Hypotenuse (force)100 lbs

RC

SC = 50

Rotational component (RC) = hypotenuse x sin

= 100 lbs x sin 50 = 76.6 lbs

Stabilizing component (SC) = hypotenuse x cos

= 100 lbs x cos 50

= 64.3 lbs

Note: 22 )3.64()6.76(100 +=

When = 30, RC = 100 lbs x sin 30 = 50 lbs; SC = 100 lbs x cos 30 = 86.6 lbs.

When = 45, RC = 100 lbs x sin 45 = 70.7 lbs; SC = 100 lbs x cos 45 = 70.7 lbs.

When = 75, RC = 100 lbs x sin 75 = 96.6 lbs; SC = 100 lbs x cos 75 = 25.88 lbs.

The larger the angle (up to 90), the greater the rotary force and the less the

stabilizing force. Note that the rotary and stabilizing forces are equal when the

angle of pull = 45.

What force must be exerted by the biceps brachii to start flexion of the forearm ( =5)? When the angle is 30, what force must be exerted?

F arm = 2 inchesR arm = 12 inches

A

F = 5

R 10 lb weight in hand

= 30 = 90

= 120

F x 2 inches = 10 lbs x 12 inchesF = 60 lbs

This (60 lbs) is the force exerted by the biceps brachii when the muscle is at a right

angle to the lever. Both angles (the insertion angle, , and its reciprocal) are 90.

At this point the force is all rotary (i.e., all 60 lbs is rotary force; RC= 60 lbs x sin 90

= 60 lbs). The muscle is most efficient at this point since no energy is beingexpended to stabilize the joint.

The force exerted by the biceps brachii at an angle of 5 with respect to the lever is:

F5 = 60 lbs sin 5 = 688 lbs

Thus, a muscle must be strong when under stretch. Nearly all of this force is

stabilizing (SC= 688 lbs x cos 5 = 685 lbs. When the angle of pull is 30, the force

exerted by the biceps brachii is:

F30 = 60 lbs sin 30 = 120 lbs

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Approximately 104 lbs (of the 120 lbs) is stabilizing and 60 lbs is rotational when the

angle of pull is 30. As the angle of pull approaches 90, more of the force

becomes rotary and less stabilizing (i.e., the effectiveness and efficiency of muscle

force increase up to 90). When the angle is between 0 and 90, the insertion

angle and angle of pull are the same. When the angle exceeds 90, the reciprocalangle becomes the angle of pull. In the fourth frame of the above diagram, the

insertion angle is 120, but the angle of pull is 60.

Lets look at the rotary component as a way of determining the best angle of pull.We can continue to use the biceps brachii and a 10 lb weight as the resistanceforce as an example. We are not trying to determine how much force is exerted bythe biceps brachii, but rather the angle at which rotary force is greatest. Thus, weonly need to calculate the rotary component using the 10 lb weight as the originalforce (FO).

RC = 10 lbs x sin 5 = 0.87 lbsRC = 10 lbs x sin 15 = 2.59 lbs

RC = 10 lbs x sin 30 = 5 lbs

RC = 10 lbs x sin 80 = 9.85 lbsRC = 10 lbs x sin 90 = 10 lbs

RC = 10 lbs x sin 120 = 8.66 lbs

These calculations reinforce the fact that a 90 angle is the most efficient and

effective angle of pull since all the force being exerted is rotary. Note that when the

angle of insertion is 120, the stabilizing component is a negative 5 lbs and is aptly

called the dislocating component.

Let us consider one more problem working with rotary and stabilizing componentsof force. Suppose the deltoid is pulling on the abducted humerus with a force of

100 lbs. The action line (hypotenuse) forms a 20 angle (angle of pull) with thehumerus. What are the rotary and stabilizing components of the 100 lb force?

100 lbs

= 20

RC = 100 lbs x sin 20 = 34.2 lbs

SC = 100 lbs x cos 20 = 93.9 lbs

Observe that the deltoid exerts predominantly a stabilizing forceuntil the distal end of the humerus is pulled well above the glenoid

cavity (i.e., when the angle of pull 45).

Consider the origin and insertion of the brachialis. It can never achieve a 90 angle relativeto the forearm. Thus, it is a good stabilizer. The biceps brachii is even more powerful whenthe humerus is extended (B) because it is stretched a little more due to its position relative

to the shoulder. This occurs even though the angle is 90 in both positions (A and B).

A

biceps brachii

B

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brachialis

Another variable to consider is the distance between the axis and the resistance (weight).Moving the weight closer to the axis will shorten the resistance arm.

Torque

If a force exerted on a body that can rotate about a pivot or fulcrum point, the forceis said to generate a torque. Thus, muscle force creates a torque. Usually thecontraction force of a muscle cannot be measured directly, so torque is usedinstead. The RC force is perpendicular to the lever arm (i.e., axis to insertion).Torque is the product of the RC force and the perpendicular distance (PD; betweenthe axis and insertion).

100 lbs

= 40

3 inches

RC = 100 lbs x sin 40 = 64.3 lbs

Torque = RC x PD = 64.3 lbs x 0.25 ft = 16.1 ft-lbs

(3 inches = 0.25 ft)

Practical applications of concepts involved with levers and muscle forceProblem 1: Ergonomics at a computer

Which of the computer operators would be most likely to complain of neck pain?

One with the head above the upper torso or one with the head leaning forwards.The weight of the operators heads is 5 kg. Operator As head is erect and operator

Bs head is in 45 of flexion. When the head is erect, the distance from the axis of

motion (C5 disc) to the c/g of the head is 2 cm. When the head is flexed 45, the

distance is 10 cm. The distance of the erector spinae to the axis of motion is 4 cm

when the head is erect and 5 cm when it is flexed 45. The speed of gravity is 10

m/sec2. What is the force of the erector spinae (and the reaction force on C5) when

the head is erect? When the head is flexed 45?

c/g

0.02 m

0.04 m50 N

25 N(erectorspinae)

5 kg x 10 m/sec2 = 50 N (force of gravity)

F x fa = R x ra

F = (50 N x 0.02 m) 0.04 m = 25 N

The force of the erector spinae is 25 N or 5.63 lbs(N x 0.225 = lbs). This is the muscle force requiredto maintain the head in an erect posture. The

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C5 (axis of motion)reaction force on C5 is 75 N (16.88 lbs), the sum ofthe force of gravity and the force of the erectorspinae.

50 N 100 N

0.05 m

0.1 m

F x fa = R x ra

F = (50 N x 0.1 m) 0.05 m = 100 N

The force of the erector spinae is 100 N or 22.5lbs. This is the muscle force required to maintain

the head in a 45 flexed position. The reaction

force on C5 is 150 N (33.75 lbs), the sum of theforce of gravity and the force of the erector spinae.

What if the person had a small neck (fa = 0.03 m)?The force of the erector spinae would be 167 N(37.6 lbs).

From these calculations, it is obvious that the flexed position puts significantly morestrain on the erector spinae (a 4-fold increase in force). The head in the erectposition is preferable for minimizing neck pain and upper back tightness.

Problem 2: Lifting boxes

Workers must lift 25 and 46 cm boxes to shoulder height. Both types of boxes

weight 18 kg (40 lbs). It is assumed that the erector spinae will counteract the

force of the boxes. The force of these muscles operates at a distance of 5 cm fromthe axis of movement at the L5 disc. Workers have an upper body weight of about

40 kg. With the box lifted to shoulder height, the c/g is 2 cm dorsal to the axis ofmovement. The 25 cm box is held so that its c/g is 20 cm from the axis of motion.The 46 cm box is held so that its c/g is 40 cm from the axis of motion. The force ofgravity acting on each box is 180 N (10 m/sec2 x 18 kg = 180 N). The upper bodycan exert a counter force of 400 N (10 m/sec2 x 40 kg) if the person leansbackwards such that the c/g falls 2 cm behind the axis of motion (L5) resulting in acounter force of 8 Nm (400 N x 0.02 m = 8 Nm).

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What force must the erector spinae muscles exert to lift the 25 cm box? the 46 cmbox?

0.2m

180 N560 N

0.02 m

L5400 N

F = (180 N x 0.2 m) 0.05 m = 720 N

When the back is straight, the erectorspinae exerts 720 N.

F = (36 Nm 8 Nm) 0.05 m = 560 N

When the person leans backward, the

erector spinae exert 560 N (126 lbs).The counter force of 8 Nm is subtractedfrom the resistance force.

0.4 m

180 N 1280 N

0.02 m

400 N

180 N x 0.4 m = 72 Nm

F= (72 Nm 8 Nm) 0.05 m = 1,280 N

The erector spinae exert 1,280 N (288lbs) when leaning backwards. Eventhough both boxes weigh the same, theresistance arm is lengthened with the

larger box because the arms must beextended more. The erector spinaemust exert more than twice the force tocarry the same weight.

Leaning backwards shifts the c/g and, thus, displaces the resistance arm to make iteasier. Bringing the load in closer to the the body also shortens the resistance arm

10

0.0

5m

0.0

5m

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and will also decrease the weight if the load can be rested on the chest orabdomen.

People with large bellies cannot bring the box/load in as close to their bodies, whichincreases the resistance arm making the load heavier. They tend to shift their c/g

back relative to L5 producing lordosis, which lengthens the force arm anddecreases the force (load) on the erector spinae.

Problem 3: Leaning over a sink

When standing (position A), the c/g of the head and trunk falls 17 cm from the axisof motion in the L5 disc, and the c/g of the arms falls 20 cm from the L5 disc. Whenbending over, the c/g of the head and trunk is 31 cm from the L5 disc, and the c/g ofthe arms is 52 cm from the L5 disc. The person weighs 60 kg. The weight of thearms is assumed to be 6 kg; the upper body is 34 kg. The erector spinae, whichcounters the force of gravity acting on the head, trunk and arms, is located 5 cm

from the axis of motion in the L5 disc. It pulls at an angle of 90 to the surface of

the vertebral body.

How much force must the erector spinae exert to counter the weight of the head,trunk and arms when standing and when bending over?

(Position A)

0.17 m340 N

1396 N 60 N0.2 m

0.05 m

m05.0

)m17.0xN340()m2.0xN60(F

+

=

F = 1396 N

(Position B)

2732 N

0.31 m0.05 m 340 N

0.52 m 60 N

m05.0

m52.0xN60()m31.0xN340(F

+=

F = 2,732 N

The resistance offered by the arms and upper body must be considered separatelybecause they have different resistance arms. What if the erector spinae was thicker(stronger) or thinner (consider aging and/or sedentary people) than 5 cm? Thecalculations will demonstrate that work is decreased as muscle thickness increases.

Problem 4: Getting off the throne

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Is it easier to get off the throne from an erect sitting posture or by leaning forwardfirst? Would armrests help someone who has weak quadriceps? For the examplebelow, the upper body weight is 48 kg. The knees are the axis of motion. It is 33cm from the c/g of gravity to the knees when sitting erect; 22 cm when leaningforward. The moment arm for the quadriceps with the knees (force arm) is 5 cm.

3168 N

0.05 m0.33 m

480 N

F = (480 N x 0.33 m) 0.05 m = 3168 N

2112 N0.22 m

480 N 0.05 m

F = (480 N x 0.22 m) 0.05 m = 2112 N

Leaning forward shortens the resistance arm as the c/g is moved forward.Consequently, leaning forward makes it easier to stand up from the throne (about1000 N/225 lbs easier). Using the arms to push off the knees or an armrest createsa reaction force that acts in the same direction as the quadriceps force, which willalso make standing easier. People with weak quadriceps should lean forward anduse armrests to stand up.

Problem 4: Squats

Performing squats can be dangerous due, in part, to excessive force acting on thelow back. Is it better to lean forward or maintain a more erect posture whenperforming squats? To answer this question, consider the illustrations below and

calculate the muscle force at the low back for each illustration.

c/g of head and trunk from axis = 0.14 mc/g of arms from axis = 0.17 m(L5 is the axis)

0.05 m

weight of head, trunk andweights = 600 N

weight of arms = 60 N

= 0.31 m 0.31 m= 0.40 m

0.05 m

= 600 N= 60 N

m05.0

)m17.0xN60()m14.0xN600(F

+=

F = 1884 N

m05.0

40.0xN60()m31.0xN600(F

+=

F = 4200 N

Leaning forward more puts much more pressure/stress on the low back.

How much force must the quadriceps exert to assume an upright posture?

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c/g of head and trunk from axis = 0.14 m(knees are the axis)weight of body and weight = 660 Nforce arm distance = 0.05 m

0.14 m

0.05 m

660 N

F = (660 N x 0.14 m) 0.05 m = 1848 N

= 0.07 m

= 660 N= 0.05 m

0.07 m0.05 m

660 N

F = (660 N x 0.07 m) 0.05 m = 924 N

We can conclude that leaning forward more hurts the low back more, but also lightens theload on the quadriceps. One may be able to lift more weight by leaning forward more, butincreases the risk of low back injury.

Problem 5: Forces acting on the neck and back

Refer to the illustration below and calculate the force of the erector spinae in theneck when the head is erect (1) and tilted (2), the force of the erector spinae in thelow back when carrying the box (3), the erector spinae in the low back when theknee is extended (4) and flexed (5).

The weight of the upper body is 40 kg (40 kg x 10m/sec2 = 400 N). Distances are as follows: a = 0.03m, b = 0.02 m, c = 0.05 m, d = 0.15 m, e = 0.2 m, f =

0.06 m, g = 0.02 m, h = 0.01 m, i = 0.9 m, j = 0.52 m.

1. F = (45 N x 0.02 m) 0.03 m = 30 N

2. F = (45 N x 0.15 m) 0.05 m = 135 N

3.m06.0

)m02.0xN400()m2.0xN180(F

=

4. F = (280 N x 0.9 m) 0.01 m = 25,200 N

5. F = (280 N x 0.52 m) 0.01 m = 14,560 N

a b 45 N

1c d

2 e180 N

f gi

3

400 N jh

4, 5

280 N 280 N

For solving the force of the erector spinae in the low back when carrying the box (3),8 Nm (400 N x 0.02 m) must be subtracted because the upper body falls behing theaxis (L5) by 2 cm. The resistance arm is shortened and the erector spinae does nothave to work as hard. Note that when the knee is flexed, the pelvic region is morestable and the erector spinae does not have to exert as much force to maintain anerect posture.

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Flexing the knees is also important when performing situps. Keeping the kneesstraight puts more force on the low back (erector spinae) producing lordosis and lowback pain. Arching shortens the force arm. Chronic shortening of a muscle leads toadaptive shortening (i.e., the muscle remains shortened lordosis). Flexing the

knees shortens the resistance arm, which reduces the force on the low back.

0.05 m 0.04 m

Selected ReferencesDyson, G.H.G. (1962). The Mechanics of Athletics. St. Paul's House, Warwick Lane,London EC4: University of London Press Ltd.Grabiner, M.D. (1993). Current Issues in Biomechanics. Champaign, IL: Human KineticsPublishers.Hall, S.J. (1999). Basic Biomechanics. Dubuque, IA: WCB McGraw-Hill.Hay, J.G. (1985). The Biomechanics of Sports Techniques. Englewood Cliffs, NJ:Prentice-Hall, Inc.LeVeau, B. (1977). Williams and Lissner: Biomechanics of Human Motion. Philadelphia,PA: W.B. Saunders Company.Northrip, J.W., Logan, G.A., and McKinney, W.C. (1974). Introduction to BiomechanicAnalysis of Sport. Dubuque, Iowa: Wm. C. Brown Company Publishers.Simonian, C. (1981). Fundamentals of Sports Biomechanics. Englewood Cliffs, NJ:Prentice-Hall, Inc.Williams, M. and Lissner, H.R. (1962). Biomechanics of Human Motion. Philadelphia, PA:W.B. Saunders Company.

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