Fundamentals of biomechanics

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Fundamentals of biomechanics

  1. 1. Applications of Statics to Biomechanics55.1 Skeletal JointsThe human body is rigid in the sense that it can maintain aposture, and flexible in the sense that it can change itsposture and move. The flexibility of the human body is dueprimarily to the joints, or articulations, of the skeletal sys-tem. The primary function of joints is to provide mobility tothe musculoskeletal system. In addition to providing mobil-ity, a joint must also possess a degree of stability. Sincedifferent joints have different functions, they possess vary-ing degrees of mobility and stability. Some joints areconstructed so as to provide optimum mobility. For example,the construction of the shoulder joint (ball-and-socket)enables the arm to move in all three planes (triaxial motion).However, this high level of mobility is achieved at theexpense of reduced stability, increasing the vulnerability ofthe joint to injuries, such as dislocations. On the other hand,the elbow joint provides movement primarily in one plane(uniaxial motion), but is more stable and less prone toinjuries than the shoulder joint. The extreme case ofincreased stability is achieved at joints that permit no rela-tive motion between the bones constituting the joint. Thecontacting surfaces of the bones in the skull are typicalexamples of such joints.The joints of the human skeletal system may be classifiedbased on their structure and/or function. Synarthrodial joints,such as those in the skull, are formed by two tightly fittingbones and do not allow any relative motion of the bonesforming them. Amphiarthrodial joints, such as those betweenthe vertebrae, allow slight relative motions, and feature anintervening substance (a cartilaginous or ligamentous tissue)whose presence eliminates direct bone-to-bone contact. Thethird and mechanically most significant type of articulationsare called diarthrodial joints which permit varying degrees ofrelative motion and have articular cavities, ligamentouscapsules, synovial membranes, and synovial fluid (Fig. 5.1).The articular cavity is the space between the articulatingbones. The ligamentous capsule holds the articulating bonestogether. The synovial membrane is the internal liningof the ligamentous capsule enclosing the synovial fluidwhich serves as a lubricant. The synovial fluid is a viscousmaterial which functions to reduce friction, reduce wear andtear of the articulating surfaces by limiting direct contactbetween them, and nourish the articular cartilage liningthe surfaces. The articular cartilage, on the other hand, is aspecialized tissue designed to increase load distribution onthe joints and provide a wear-resistant surface that absorbsshock. Various diarthrodial joints can be further categorizedas gliding (e.g., vertebral facets), hinge (elbow and ankle),pivot (proximal radioulnar), condyloid (wrist), saddle(carpometacarpal of thumb), and ball-and-socket (shoulderand hip).The nature of motion about a diarthrodial joint and thestability of the joint are dependent upon many factors,including the manner in which the articulating surfaces fittogether, the properties of the joint capsule, the structure andlength of the ligaments around the joint, and the number andorientation of the muscles crossing the joint.Fig. 5.1 A diarthrodial joint: (1) Bone, (2) ligamentous capsule, (3, 4)synovial membrane and fluid, (5, 6) articular cartilage and cavityN. O¨ zkaya et al., Fundamentals of Biomechanics: Equilibrium, Motion, and Deformation,DOI 10.1007/978-1-4614-1150-5_5, # Springer Science+Business Media, LLC 201261
  2. 2. 5.2 Skeletal MusclesIn general, there are over 600 muscles in the human body,accounting for about 45of the total body weight.There are three types of muscles: cardiac, smooth, andskeletal. Cardiac muscle is the contractive tissue found inthe heart that pumps the blood for circulation. Smooth mus-cle is found in the stomach, intestinal tracts, and the walls ofblood vessels. Skeletal muscle is connected to the bones ofthe body and when contracted, causes body segments tomove.Movement of human body segments is achieved as aresult of forces generated by skeletal muscles that convertchemical energy into mechanical work. The structuralunit of skeletal muscle is the muscle fiber, which is com-posed of myofibrils. Myofibrils are made up of actin andmyosin filaments. Muscles exhibit viscoelastic materialbehavior. That is, they have both solid and fluid-likematerial properties. Muscles are elastic in the sense thatwhen a muscle is stretched and released it will resumeits original (unstretched) size and shape. Muscles are vis-cous in the sense that there is an internal resistance tomotion.A skeletal muscle is attached, via soft tissues such asaponeuroses and/or tendons, to at least two different bonescontrolling the relative motion of one segment with respectto the other. When its fibers contract under the stimulation ofa nerve, the muscle exerts a pulling effect on the bones towhich it is attached. Contraction is a unique property of themuscle tissue. In engineering mechanics, contraction impliesshortening under compressive forces. In muscle mechanics,contraction can occur as a result of muscle shortening ormuscle lengthening, or it can occur without any change inthe muscle length. Furthermore, the result of a muscle con-traction is always tension: a muscle can only exert a pull.Muscles cannot exert a push.There are various types of muscle contractions: a concen-tric contraction occurs simultaneously as the length ofthe muscle decreases (e.g., the biceps during flexion of theforearm); a static contraction occurs while muscle lengthremains constant (the biceps when the forearm is flexed andheld without any movement); and an eccentric contractionoccurs as the length of the muscle increases (the bicepsduring the extension of the forearm). A muscle can causemovement only while its length is shortening (concentriccontraction). If the length of a muscle increases during aparticular activity, then the tension generated by the musclecontraction is aimed at controlling the movement of thebody segments associated with that muscle (excentric con-traction). If a muscle contracts but there is no segmentalmotion, then the tension in the muscle balances the effectsof applied forces such as those due to gravity (isometriccontraction).The skeletal muscles can also be named according to thefunctions they serve during a particular activity. For exam-ple, a muscle is called agonist if it causes movement throughthe process of its own contraction. Agonist muscles are theprimary muscles responsible for generating a specific move-ment. An antagonist muscle opposes the action of anothermuscle. Synergic muscle is that which assists the agonistmuscle in performing the same joint motion.5.3 Basic ConsiderationsIn this chapter, we want to apply the principles of staticsto investigate the forces involved in various musclegroups and joints for various postural positions of thehuman body and its segments. Our immediate purpose is toprovide answers to questions such as: what tension mustthe neck extensor muscles exert on the head to supportthe head in a specified position? When a person bends,what would be the force exerted by the erector spinae onthe fifth lumbar vertebra? How does the compression atthe elbow, knee, and ankle joints vary with externallyapplied forces and with different segmental arrangements?How does the force on the femoral head vary with loadscarried in the hand? What are the forces involved in variousmuscle groups and joints during different exerciseconditions?The forces involved in the human body can be grouped asinternal and external. Internal forces are those associatedwith muscles, ligaments, and tendons, and at the joints.Externally applied forces include the effect of gravitationalacceleration on the body or on its segments, manually and/ormechanically applied forces on the body during exercise andstretching, and forces applied to the body by prostheses andimplements. In general, the unknowns in static problemsinvolving the musculoskeletal system are the joint reactionforces and muscle tensions. Mechanical analysis of a jointrequires that we know the vector characteristics of tensionin the muscle including the proper locations of muscleattachments, the weights or masses of body segments, thecenters of gravity of the body segments, and the anatomicalaxis of rotation of the joint.5.4 Basic Assumptions and LimitationsThe complete analysis of muscle forces required to sustainvarious postural positions is difficult because of the complexarrangement of muscles within the human body and becauseof limited information. In general, the relative motion of62 5 Applications of Statics to Biomechanics
  3. 3. body segments about a given joint is controlled by more thanone muscle group. To be able to reduce a specific problem ofbiomechanics to one that is statically determinate and applythe equations of equilibrium, only the muscle group that isthe primary source of control over the joint can be taken intoconsideration. Possible contributions of other muscle groupsto the load-bearing mechanism of the joint must be ignored.Note however that approximations of the effect of othermuscles may be made by considering their cross-sectionalareas and their relative positions in relation to the joint. Also,if the phasic activity of muscles is known via someexperiments such as the electromyography (EMG)measurements of muscle signals, then the tension in differ-ent muscle groups may be estimated.To apply the principles of statics to analyze the mechan-ics of human joints, we shall adopt the followingassumptions and limitations:• The anatomical axes of rotation of joints are known.• The locations of muscle attachments are known.• The line of action of muscle tension is known.• Segmental weights and their centers of gravity areknown.• Frictional factors at the joints are negligible.• Dynamic aspects of the problems will be ignored.• Only two-dimensional problems will be considered.These analyses require that the anthropometric data aboutthe segment to be analyzed must be available. For thispurpose, there are tables listing anthropometric informationincluding average weights, lengths, and centers of gravity ofbody segments. See Chaffin et al. (1999), Roebuck (1995),and Winter (2004) for a review of the anthropometric dataavailable.It is clear from this discussion that we shall analyzecertain idealized problems of biomechanics. Based on theresults obtained and experience gained, these models may beexpanded by taking additional factors into consideration.However, a given problem will become more complex asmore factors are considered.In the following sections, the principles of statics areapplied to analyze forces involved at and around the majorjoints of the human body. First, a brief functional anatomy ofeach joint and related muscles is provided, and specific bio-mechanical problems are constructed. For a more completediscussion about the functional anatomy of joints, see textssuch as Nordin and Frankel (2012) and Thompson (1989).Next, an analogy is formed between muscles, bones, andhuman joints, and certain mechanical elements such ascables, beams, and mechanical joints. This enables us toconstruct a mechanical model of the biologicalsystem under consideration. Finally, the procedure outlinedin section “Procedure to Analyze Systems in Equilibrium”of Chap. 4 is applied to analyze the mechanical modelthus constructed. See LeVeau (1992) for additional examplesof the application of the principles of statics to biomechanics.5.5 Mechanics of the ElbowThe elbow joint is composed of three separate articulations(Fig. 5.2). The humeroulnar joint is a hinge (ginglymus)joint formed by the articulation between the spool-shapedtrochlea of the distal humerus and the concave trochlearfossa of the proximal ulna. The structure of the humeroulnarjoint is such that it allows only uniaxial rotations, confiningthe movements about the elbow joint to flexion (movementof the forearm toward the upper arm) and extension (move-ment of the forearm away from the upper arm). Thehumeroradial joint is also a hinge joint formed between thecapitulum of the distal humerus and the head of the radius.The proximal radioulnar joint is a pivot joint formed by thehead of the radius and the radial notch of the proximal ulna.This articulation allows the radius and ulna to undergorelative rotation about the longitudinal axis of one or theother bone, giving rise to pronation (the movement experi-enced while going from the palm-up to the palm-down) orsupination (the movement experienced while going from thepalm-down to the palm-up).The muscles coordinating and controlling the movementof the elbow joint are illustrated in Fig. 5.3. The bicepsbrachii muscle is a powerful flexor of the elbow joint, par-ticularly when the elbow joint is in a supinated position. It isthe most powerful supinator of the forearm. On the distalside, the biceps is attached to the tuberosity of the radius, andon the proximal side, it has attachments at the top of thecoracoids process and upper lip of the glenoid fossa. Anotherimportant flexor is the brachialis muscle which, regardless offorearm orientation, has the ability to produce elbow flexion.Fig. 5.2 Bones of the elbow:(1) humerus, (2) capitulum,(3) trochlea, (4) radius, (5) ulna5.5 Mechanics of the Elbow 63
  4. 4. It is then the strongest flexor of the elbow. It has attachmentsat the lower half of the anterior portion of the humerus andthe coronoid process of the ulna. Since it does not inset onthe radius, it cannot participate in pronation or supination.The most important muscle controlling the extensionmovement of the elbow is the triceps brachii muscle. It hasattachments at the lower head of the glenoid cavity of thescapula, the upper half of the posterior surface of thehumerus, the lower two thirds of the posterior surface ofthe humerus, and the olecranon process of the ulna. Prona-tion and supination movements of the forearm are performedby the pronator teres and supinator muscles, respectively.The pronator teres is attached to the lower part of the innercondyloid ridge of the humerus, the medial side of theulna, and the middle third of the outer surface of the radius.The supinator muscle has attachments at the outer condyloidridge of the humerus, the neighboring part of the ulna and theouter surface of the upper third of the radius.Common injuries of the elbow include fractures anddislocations. Fractures usually occur at the epicondylesof the humerus and the olecranon process of the ulna.Another group of elbow injuries are associated with overuse,which causes an inflammatory process of the tendons of anelbow that has been damaged by repetitive motions. Theseinclude tennis elbow and golfer’s elbow syndromes.Example 5.1 Consider the arm shown in Fig. 5.4. The elbowis flexed to a right angle and an object is held in the hand.The forces acting on the forearm are shown in Fig. 5.5a, andthe free-body diagram of the forearm is shown on amechanical model in Fig. 5.5b. This model assumes thatthe biceps is the major flexor and that the line of action ofthe tension (line of pull) in the biceps is vertical.Point O designates the axis of rotation of the elbow joint,which is assumed to be fixed for practical purposes. Point Ais the attachment of the biceps muscle on the radius, point Bis the center of gravity of the forearm, and point C is a pointon the forearm that lies along a vertical line passing throughthe center of gravity of the weight in the hand. The distancesbetween point O and points A, B, and C are measured as a, b,and c, respectively. Wo is the weight of the object held in thehand and W is the total weight of the forearm. FM is themagnitude of the force exerted by the biceps on the radius,and FJ is the magnitude of the reaction force at the elbowjoint. Notice that the line of action of the muscle force isassumed to be vertical. The gravitational forces are verticalas well. Therefore, for the equilibrium of the lower arm, theline of action of the joint reaction force must also be vertical(a parallel force system).Fig. 5.3 Muscles of the elbow: (1) biceps, (2) brachioradialis, (3)brachialis, (4) pronator teres, (5) triceps brachii, (6) anconeus, (7)supinatorFig. 5.4 Example 5.1Fig. 5.5 Forces acting on the lower arm64 5 Applications of Statics to Biomechanics
  5. 5. The task in this example is to determine the magnitudes ofthe muscle tension and the joint reaction force at the elbow.SolutionWe have a parallel force system, and the unknowns are themagnitudes FM and FJ of the muscle and joint reactionforces. Considering the rotational equilibrium of the forearmabout the elbow joint and assuming the (cw) direction ispositive:XMo ¼ 0:That is, cWo þ bW À aFM ¼ 0:Then FM ¼1aðbW þ cWoÞ: (i)For the translational equilibrium of the forearm in the ydirection,XFy ¼ 0:That is, À FJ þ FM À W À Wo ¼ 0:Then FJ ¼ FM À W À Wo: (ii)For given values of geometric parameters a, b, and c, andweights W and Wo, Eqs. (i) and (ii) can be solved for themagnitudes of the muscle and joint reaction forces. Forexample, assume that these parameters are given asfollows: a ¼ 4cm; b ¼ 15cm; c ¼ 35cm; W ¼ 20N,and Wo ¼ 80N. Then from Eqs. (i) and (ii):FM ¼10:04ð0:15Þð20Þ þ ð0:35Þð80Þ½ Š ¼ 775 N ðþyÞ:FJ ¼ 775 À 20 À 80 ¼ 675 N ðÀyÞ:Remarks• The numerical results indicate that the force exerted bythe biceps muscle is about ten times larger than theweight of the object held in the position considered.Relative to the axis of the elbow joint, the length a ofthe lever arm of the muscle force is much smaller than thelength c of the lever arm enjoyed by the load held in thehand. The smaller the lever arm, the greater the muscletension required to balance the clockwise rotational effectof the load about the elbow joint. Therefore, duringlifting, it is disadvantageous to have a muscle attachmentclose to the elbow joint. However, the closer the muscle isto the joint, the larger the range of motion of elbowflexion–extension, and the faster the distal end (hand) ofthe forearm can reach its goal of moving toward the upperarm or the shoulder.• The angle between the line of action of the muscle forceand the long axis of the bone upon which the muscle forceis exerted is called the angle of pull and it is critical indetermining the effectiveness of the muscle force. Whenthe lower arm is flexed to a right angle, the muscle tensionhas only a rotational effect on the forearm about theelbow joint, because the line of action of the muscleforce is at a right angle with the longitudinal axis ofthe forearm. For other flexed positions of the forearm,the muscle force can have a translational (stabilizing orsliding) component as well as a rotational component.Assume that the linkage system shown in Fig. 5.6aillustrates the position of the forearm relative to the upperarm. n designates a direction perpendicular (normal) tothe long axis of the forearm and t is tangent to it. Assum-ing that the line of action of the muscle force remainsparallel to the long axis of the humerus, FM can bedecomposed into its rectangular components FMn andFMt. In this case, FMn is the rotational (rotatory) compo-nent of the muscle force because its primary function is torotate the forearm about the elbow joint. The tangentialcomponent FMt of the muscle force acts to compress theelbow joint and is called the stabilizing component of theFig. 5.6 Rotational (FMn) and stabilizing or sliding (FMt) componentsof the muscle force5.5 Mechanics of the Elbow 65
  6. 6. muscle force. As the angle of pull approaches 90, themagnitude of the rotational component of the muscleforce increases while its stabilizing component decreases,and less and less energy is “wasted” to compress theelbow joint. As illustrated in Fig. 5.6b, the stabilizingrole of FMt changes into a sliding or dislocating rolewhen the angle between the long axes of the forearmand upper arm becomes less than 90.• The elbow is a diarthrodial (synovial) joint. A ligamentouscapsule encloses an articular cavity which is filled withsynovial fluid. Synovial fluid is a viscous material whoseprimary function is to lubricate the articulating surfaces,thereby reducing the frictional forces that may developwhile one articulating surface slides over the other. Thesynovial fluid also nourishes the articulating cartilages. Acommonpropertyoffluidsisthattheyexert pressures(forceper unit area) that are distributed over the surfaces theytouch. The fluid pressure always acts in a direction towardand perpendicular to the surface it touches having a com-pressiveeffectonthe surface.Notethat inFig.5.7, the smallvectors indicating the fluid pressure have components in thehorizontal and vertical directions. We determined that thejoint reaction force at the elbow acts vertically downwardon the ulna. This implies that the horizontal components ofthese vectors cancel out (i.e., half pointing to the left andhalf pointing to the right), but their vertical components (onthe ulna, almost all of them are pointing downward) add upto form the resultant force FJ (shown with a dashed arrow inFig.5.7c).Therefore,thejointreactionforceFJ correspondsto the resultant of the distributed force system (pressure)applied through the synovial fluid.• The most critical simplification made in this example isthat the biceps was assumed to be the single muscle groupresponsible for maintaining the flexed configuration ofthe forearm. The reason for making such an assumptionwas to reduce the system under consideration to one thatis statically determinate. In reality, in addition to thebiceps, the brachialis and the brachioradialis are primaryelbow flexor muscles.Consider the flexed position of the arm shown inFig. 5.8a. The free-body diagram of the forearm is shownin Fig. 5.8b. FM1, FM2, and FM3 are the magnitudes of theforces exerted on the forearm by the biceps, the brachialis,and the brachioradialis muscles with attachments at pointsA1, A2, and A3, respectively. Let y1, y2, and y3 be the anglesthat the biceps, the brachialis, and the brachioradialismuscles make with the long axis of the lower arm. Ascompared to the single-muscle system which consisted oftwo unknowns (FM and Fj), the analysis of this three-musclesystem is quite complex. First of all, this is not a simpleparallel force system. Even if we assume that the locations ofmuscle attachments (A1,A2, and A3), their angles of pull (y1,y2, and y3) and the lengths of their moment arms (a1, a2, anda3) as measured from the elbow joint are known, there arestill five unknowns in the problem (FM1, FM2, FM3, Fj, and b,where the angle b is an angle between Fj and the long axes ofthe forearm). The total number of equations available fromstatics is three:XMo ¼ 0 : a1FM1 þa2FM2 þa3FM3 ¼ bW þcWo: (iii)Fig. 5.8 Three-muscle systemFig. 5.7 Explaining the joint reaction force at the elbow66 5 Applications of Statics to Biomechanics
  7. 7. XFx ¼ 0 : FJx ¼ FM1x þ FM2x þ FM3x: (iv)XFy ¼ 0 : FJy ¼ FM1y þ FM2y þ FM3y À W À Wo: (v)Note that once the muscle forces are determined, Eqs. (iv)and (v) will yield the components of the joint reaction forceFJ. As far as the muscle forces are concerned, we have onlyEq. (iii) with three unknowns. In other words, we have astatically indeterminate problem. To obtain a unique solution,we need additional information relating FM1, FM2, and FM3.There may be several approaches to the solution of thisproblem. The criteria for estimating the force distributionamong different muscle groups may be established by (1)using cross-sectional areas of muscles, (2) using EMGmeasurements of muscle signals, and (3) applying certainoptimization techniques. It may be assumed that each mus-cle exerts a force proportional to its cross-sectional area.If S1, S2, and S3 are the cross-sectional areas of the biceps,the brachialis, and the brachioradialis, then this criteria maybe applied by expressing muscle forces in the followingmanner:FM2 ¼ k21 FM1 with k21 ¼S2S1: (vi)FM3 ¼ k31 FM1 with k31 ¼S3S1: (vii)If constants k21 and k31 are known, then Eqs. (vi) and (vii)can be substituted into Eq. (iii), which can then be solved forFM1:FM1 ¼b W þ c Woa1 þ a2 k21 þ a3 k31:Substituting FM1 back into Eqs. (vi) and (vii) will thenyield the magnitudes of the forces in the brachialis and thebrachioradialis muscles. The values of k21 and k31 may also beestimated by using the amplitudes of muscle EMG signals.This statically indeterminate problem may also be solvedby considering some optimization techniques. If the purposeis to accomplish a certain task (static or dynamic) in the mostefficient manner, then the muscles of the body must act tominimize the forces exerted, the moments about the joints (fordynamic situations), and/or the work done by the muscles.The question is, what force distribution among the variousmuscles facilitates the maximum efficiency?5.6 Mechanics of the ShoulderThe bony structure and the muscles of the shoulder complexare illustrated in Figs. 5.9 and 5.10. The shoulder forms thebase for all upper extremity movements. The complexstructure of the shoulder can be divided into two: the shoul-der joint and the shoulder girdle.The shoulder joint, also known as the glenohumeralarticulation, is a ball-and-socket joint between the nearlyhemispherical humeral head (ball) and the shallowly con-cave glenoid fossa (socket) of the scapula. The shallownessof the glenoid fossa allows a significant freedom of move-ment of the humeral head on the articulating surface of theglenoid. The movements allowed are in the sagittal plane,flexion (movement of the humerus to the front—a forwardupward movement) and extension (return from flexion); inFig. 5.9 The shoulder: (1) sternoclavicular joint, (2) sternum, (3)glenohumeral joint, (4) clavicle, (5) acromioclavicular joint, (6)acromion process, (7) glenoid fossa, (8) scapula, (9) humerusFig. 5.10 Shoulder muscles: (1) deltoideus, (2) pectoralis minor, (3)subscapularis, (4) pectoralis major, (5) trapezius, (6) infraspinatus andteres minor, (7) latissimus dorsi, (8) levator scapulae, (9) supraspinatus,(10) rhomboideus, (11) teres major5.6 Mechanics of the Shoulder 67
  8. 8. the coronal plane, abduction (horizontal upward movementof the humerus to the side) and adduction (return fromabduction); and in the transverse plane, outward rotation(movement of the humerus around its long axis to the lateralside) and inward rotation (return from outward rotation). Theconfiguration of the articulating surfaces of the shoulderjoint also makes the joint more susceptible to instabilityand injury, such as dislocation. The stability of the joint isprovided by the glenohumeral and coracohumeral ligaments,and by the muscles crossing the joint. The major muscles ofthe shoulder joint are deltoideus, supraspinatus, pectoralismajor, coracobrachialis, latissimus dorsi, teres major, teresminor, infraspinatus, and subscapularis.The bony structure of the shoulder girdle consists ofthe clavicle (collarbone) and the scapula (shoulder blade).The acromioclavicular joint is a small synovial articulationbetween the distal clavicle and the acromion process of thescapula. The stability of this joint is reinforced by thecoracoclavicular ligaments. The sternoclavicular joint isthe articulation between the manubrium of the sternum andthe proximal clavicle. The stability of this joint is enhancedby the costoclavicular ligament. The acromioclavicular jointand the sternoclavicular joint both have layers of cartilage,called menisci, interposed between their bony surfaces.There are four pairs of scapular movements: elevation(movement of the scapula in the frontal plane) and depres-sion (return from elevation), upward rotation (turning theglenoid fossa upward and the lower medial border of thescapula away from the spinal column) and downward rota-tion (return from upward rotation), protraction (movementof the distal end of the clavicle forward) and retraction(return from protraction), and forward and backward rota-tion (rotation of the scapula about the shaft of the clavicle).Some of the main muscles that control and coordinate thesemovements are the trapezius, levator scapulae, rhomboid,pectoralis minor, serratus anterior, and subclavius.Example 5.2 Consider a person strengthening the shouldermuscles by means of dumbbell exercises. Figure 5.11illustrates the position of the left arm when the arm is fullyabducted to horizontal. The free-body diagram of the arm isshown in Fig. 5.12 along with a mechanical model of thearm. Also in Fig. 5.12, the forces acting on the arm areresolved into their rectangular components along the hori-zontal and vertical directions. Point O corresponds tothe axis of rotation of the shoulder joint, point A is wherethe deltoid muscle is attached to the humerus, point B is thecenter of gravity of the entire arm, and point C is the centerof gravity of the dumbbell. W is the weight of the arm, Wo isthe weight of the dumbbell, FM is the magnitude of thetension in the deltoid muscle, and Fj is the joint reactionforce at the shoulder. The resultant of the deltoid muscleforce makes an angle y with the horizontal. The distancesbetween point O and points A, B, and C are measured as a, b,and c, respectively.Determine the magnitude FM of the force exerted by thedeltoid muscle to hold the arm at the position shown. Alsodetermine the magnitude and direction of the reaction forceat the shoulder joint in terms of specified parameters.SolutionWith respect to the xy coordinate frame, the muscle and jointreaction forces have two components while the weights ofthe arm and the dumbbell act in the negative y direction. Thecomponents of the muscle force areFMx ¼ FM cos y ðÀxÞ: (i)FMy ¼ FM sin y ðþyÞ: (ii)Fig. 5.11 The arm is abducted to horizontalFig. 5.12 Forces acting on the arm and a mechanical modelrepresenting the arm68 5 Applications of Statics to Biomechanics
  9. 9. Components of the joint reaction force areFJx ¼ FJ cos b ðþxÞ: (iii)FJy ¼ FJ sin b ðÀyÞ (iv)b is the angle that the joint reaction force makes with thehorizontal. The line of action and direction (in terms of y)of the force exerted by the muscle on the arm are known.However, the magnitude FM of the muscle force, the magni-tude Fj, and the direction (b) of the joint reaction forceare unknowns. We have a total of three unknowns, FM, Fj,and b (or FM, FJx, and FJy). To be able to solve this two-dimensional problem, we have to utilize all three equilib-rium equations.First, consider the rotational equilibrium of the arm aboutthe shoulder joint at point O. The joint reaction forceproduces no torque about point O because its line of actionpasses through it. For practical purposes, we can neglect thepossible contribution of the horizontal component of themuscle force to the moment generated about point O byassuming that its line of action also passes through pointO. Note that this is not a critical or necessary assumption tosolve this problem. If we knew the length of its moment arm(i.e., the vertical distance between O and A), we could easilyincorporate the torque generated by FMx about point O intothe analysis. Under these considerations, there are only threemoment producing forces about point O. For the rotationalequilibrium of the arm, the net moment about point O mustbe equal to zero. Taking counterclockwise moments to bepositive:XMo ¼ 0 : a FMy À b W À c Wo ¼ 0FMy ¼1aðb W þ c WoÞ: (v)For given a; b; c; W, and Wo, Eq. (v) can be used todetermine the vertical component of the force exerted by thedeltoid muscle. Equation, (ii) can now be used to determinethe total force exerted by the muscle:FM ¼FMysin y: (vi)Knowing FM, Eq. (i) will yield the horizontal componentof the tension in the muscle:FMx ¼ FM cos y: (vii)The components of the joint reaction force can be deter-mined by considering the translational equilibrium of thearm in the horizontal and vertical directions:XFx ¼ 0 thatis, FJx ÀFMx ¼ 0; then FJx ¼ FMx: (viii)XFy ¼ 0 that is, À FJy þ FMy À W À Wo ¼ 0;then FJy ¼ FMy À W À Wo: (ix)Knowing the rectangular components of the joint reactionforce enables us to compute the magnitude of the force itselfand the angle its line of action makes with the horizontal:FJ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðFJxÞ2þ ðFJyÞ2q: (x)b ¼ tanÀ1 FJyFJx : (xi)Now consider that a ¼ 15 cm; b ¼ 30 cm;c ¼ 60 cm; y ¼ 15; W ¼ 40 N, and Wo ¼ 60 N. Then,FMy ¼10:15ð0:30Þð40Þ þ ð0:60Þð60Þ½ Š ¼ 320 N ðþyÞ:FM ¼320sin 15¼ 1;236 N:FMx ¼ ð1;236Þ ðcos 15Þ ¼ 1;194 N ðÀxÞ:FJx ¼ 1;194 N ðþxÞ:FJy ¼ 320 À 40 À 60 ¼ 220 N ðÀyÞ:FJ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1;1942þ 2202q¼ 1;214 N:b ¼ tanÀ1 2201;194 ¼ 10:Remarks• FMx is the stabilizing component and FMy is the rotationalcomponent of the deltoid muscle. FMx is approximatelyfour times larger than FMy. A large stabilizing componentsuggests that the horizontal position of the arm is notstable, and that the muscle needs to exert a high horizon-tal force to stabilize it.• The human shoulder is very susceptible to injuries. Themost common injuries are dislocations of the shoulderjoint and the fracture of the humerus. Since the socket ofthe glenohumeral joint is shallow, the head of thehumerus is relatively free to rotate about the articulatingsurface of the glenoid fossa. This freedom of movementis achieved, however, by reduced joint stability. Thehumeral head may be displaced in various ways,depending on the strength or weakness of the muscularand ligamentous structure of the shoulder, and depending5.6 Mechanics of the Shoulder 69
  10. 10. on the physical activity. Humeral fractures are anothercommon type of injuries. The humerus is particularlyvulnerable to injuries because of its unprotectedconfiguration.• Average ranges of motion of the arm about the shoulderjoint are 230during flexion–extension, and 170in bothabduction–adduction and inward–outward rotation.5.7 Mechanics of the Spinal ColumnThe human spinal column is the most complex part of thehuman musculoskeletal system. The principal functions ofthe spinal column are to protect the spinal cord; to supportthe head, neck, and upper extremities; to transfer loads fromthe head and trunk to the pelvis; and to permit a variety ofmovements. The spinal column consists of the cervical(neck), thoracic (chest), lumbar (lower back), sacral, andcoccygeal regions. The thoracic and lumbar sections of thespinal column make up the trunk. The sacral and coccygealregions are united with the pelvis and can be consideredparts of the pelvic girdle.The vertebral column consists of 24 intricate and com-plex vertebrae (Fig. 5.13). The articulations between thevertebrae are amphiarthrodial joints. A fibrocartilaginousdisk is interposed between each pair of vertebrae. The pri-mary functions of these intervertebral disks are to sustainloads transmitted from segments above, act as shockabsorbers, eliminate bone-to-bone contact, and reduce theeffects of impact forces by preventing direct contact betweenthe bony structures of the vertabrae. The articulations ofeach vertebra with the adjacent vertabrae permit movementin three planes, and the entire spine functions like a singleball-and-socket joint. The structure of the spine allows awide variety of movements including flexion–extension,lateral flexion, and rotation.Two particularly important joints of the spinal columnare those with the head (occiput bone of the skull) and thefirst cervical vertebrae, atlas, and the atlas and the secondvertebrae, the axis. The atlantooccipital joint is the unionbetween the first cervical vertebra (the atlas) and the occipi-tal bone of the head. This is a double condyloid joint andpermits movements of the head in the sagittal and frontalplanes. The atlantoaxial joint is the union between the atlasand the odontoid process of the head. It is a pivot joint,enabling the head to rotate in the transverse plane. Themuscle groups providing, controlling, and coordinatingthe movement of the head and the neck are the preverte-brals (anterior), hyoids (anterior), sternocleidomastoid(anterior–lateral), scalene (lateral), levator scapulae (lat-eral), suboccipitals (posterior), and spleni (posterior).The spine gains its stability from the intervertebral discsand from the surrounding ligaments and muscles (Fig. 5.14).The discs and ligaments provide intrinsic stability, and themuscles supply extrinsic support. The muscles of the spineexist in pairs. The anterior portion of the spine containsthe abdominal muscles: the rectus abdominis, transverseabdominus, external obliques, and internal obliques. Thesemuscles provide the necessary force for trunk flexion andmaintain the internal organs in proper position. There arethree layers of posterior trunk muscles: erector spinae,semispinalis, and the deep posterior spinal muscle groups.The primary function of the muscles located at the posteriorportion of the spine is to provide trunk extension. Thesemuscles also support the spine against the effects of gravity.The quadratus lumborum muscle is important in lateraltrunk flexion. It also stabilizes the pelvis and lumbar spine.Fig. 5.14 Selected muscles ofthe neck and spine: (1) splenius,(2) sternocleidomastoid,(3) hyoid, (4) levator scapula,(5) erector spinae, (6) obliques,(7) rectus abdominis,(8) transversus abdominisFig. 5.13 The spinal column:(1) cervical vertebrae, (2) thoracicvertebrae, (3) lumbar vertabrae,(4) sacrum70 5 Applications of Statics to Biomechanics
  11. 11. The lateral flexion of the trunk results from the actions of theabdominal and posterior muscles. The rotational movementof the trunk is controlled by the simultaneous action ofanterior and posterior muscles.The spinal column is vulnerable to various injuries. Themost severe injury involves the spinal cord, which isimmersed in fluid and protected by the bony structure.Other critical injuries include fractured vertebrae andherniated intervertebral disks. Lower back pain may alsoresult from strains in the lower regions of the spine.Example 5.3 Consider the position of the head and the neckshown in Fig. 5.15. Also shown are the forces acting on thehead. The head weighs W ¼ 50 N and its center of gravity islocated at point C. FM is the magnitude of the resultant forceexerted by the neck extensor muscles, which is applied onthe skull at point A. The atlantooccipital joint center islocated at point B. For this flexed position of the head, it isestimated that the line of action of the neck muscle forcemakes an angle y ¼ 30and the line of action of the jointreaction force makes an angle b ¼ 60with the horizontal.What tension must the neck extensor muscles exert tosupport the head? What is the compressive force applied onthe first cervical vertebra at the atlantooccipital joint?SolutionWe have a three-force system with two unknowns:magnitudes FM and Fj of the muscle and joint reaction forces.Since the problem has a relatively complicated geometry, it isconvenient to utilize the condition that for a body to be inequilibrium the force system acting on it must be eitherconcurrent or parallel. In this case, it is clear that the forcesinvolved do not form a parallel force system. Therefore, thesystem of forces under consideration must be concurrent.Recall that a system of forces is concurrent if the lines ofaction of all forces have a common point of intersection.In Fig. 5.15, the lines of action of all three forces actingon the head are extended to meet at point O. In Fig. 5.16,the forces W, FM, and FJ acting on the skull are translated topoint O, which is also chosen to be the origin of the xycoordinate frame. The rectangular components of the muscleand joint reaction forces in the x and y directions areFMx ¼ FM cos y: (i)FMy ¼ FM sin y: (ii)FJx ¼ FJ cos b: (iii)FJy ¼ FJ sin b: (iv)The translational equilibrium conditions in the x and ydirections will yieldXFx ¼ 0 that is : ÀFJx þ FMx ¼ 0;then FJx ¼ FMx: (v)XFy ¼ 0 that is : ÀW À FMy þ FJy ¼ 0;then FJy ¼ W þ FMy: (vi)Substitute Eqs. (i) and (iii) into Eq. (v):FJ cos b ¼ FM cos y: (vii)Substitute Eqs. (ii) and (iv) into Eq. (vi):FJ sin b ¼ W þ FM sin y (viii)Substitute this equation into Eq. (viii), that is,FM  cos ycos b sin b ¼ W þ FM sin y;Fig. 5.15 Forces on the skull form a concurrent systemFig. 5.16 Components of the forces acting on the head5.7 Mechanics of the Spinal Column 71
  12. 12. FM  cos y tan b ¼ W þ FM sin y;thentan b ¼W þ FM sin yFM cos y: (ix)Equation (ix) can now be solved for the unknown muscleforce FM:FM cos y tan b ¼ W þ FM sin yFM ðcos y tan b À sin yÞ ¼ WFM ¼Wcos y tan b À sin y: (x)Equation (x) gives the tension in the muscle as a functionof the weight W of the head and the angles y and b that thelines of action of the muscle and joint reaction forces makewith the horizontal. Substituting the numerical values of W,y, and b will yieldFM ¼50ðcos 30Þ ðtan 60Þ À ð sin 30Þ¼ 50 N:From Eqs. (i) and (ii),FMx ¼ ð50Þðcos 30Þ ¼ 43 N ðþxÞFMy ¼ ð50Þðsin 30Þ ¼ 25 N ðÀyÞ:From Eqs. (v) and (vi),FJx ¼ 43 N ðÀxÞFJy ¼ 50 þ 25 ¼ 75 N ðþyÞ:The resultant of the joint reaction force can be computedeither from Eqs. (iii) or (iv). Using Eq. (iii)FJ ¼FJxcos b¼43cos 60¼ 86 N:Remarks• The extensor muscles of the head must apply a force of50 N to support the head in the position considered. Thereaction force developed at the atlantooccipital joint isabout 86 N.• The joint reaction force can be resolved into two rectan-gular components, as shown in Fig. 5.17. FJn is themagnitude of the normal component of FJ compressingthe articulating joint surface, and FJt is the magnitude ofits tangential component having a shearing effect on thejoint surfaces. Forces in the muscles and ligaments ofthe neck operate in a manner to counterbalance thisshearing effect.Example 5.4 Consider the weight lifter illustrated inFig. 5.18, who is bent forward and lifting a weight Wo.At the position shown, the athlete’s trunk is flexed by anangle y as measured from the upright (vertical) position.The forces acting on the lower portion of the athlete’sbody are shown in Fig. 5.19 by considering a section passingthrough the fifth lumbar vertebra. A mechanical model of theathlete’s lower body (the pelvis and legs) is illustrated inFig. 5.20 along with the geometric parameters of the prob-lem under consideration. W is the total weight of the athlete,W1 is the weight of the legs including the pelvis, ðW þ W0Þis the total ground reaction force applied to the athletethrough the feet (at point C), FM is the magnitude of theresultant force exerted by the erector spinae musclessupporting the trunk, and FJ is the magnitude of the com-pressive force generated at the union (point O) of the sacrumFig. 5.17 Normal and shear components of the joint reaction forceFig. 5.18 A weight lifter72 5 Applications of Statics to Biomechanics
  13. 13. and the fifth lumbar vertebra. The center of gravity of thelegs including the pelvis is located at point B. Relative topoint O, the lengths of the lever arms of the muscle force,lower body weight, and ground reaction force are measuredas a; b, and c, respectively.Assuming that the line of pull of the resultant muscleforce exerted by the erector spinae muscles is parallel tothe trunk (i.e., making an angle y with the vertical), deter-mine FM and FJ in terms of, b; c; y; W0; W1, and W.SolutionIn this case, there are three unknowns: FM, FJx, and FJy. Thelengths of the lever arms of the muscle force, ground reactionforce, andthe gravitational force of the legs including the pelvisare given as measured from point O. Therefore, we can applythe rotational equilibrium condition about point O to determinethe magnitude FM of the resultant force exerted by the erectorspinae muscles. Considering clockwise moments to be positiveXMo ¼ 0 : aFM þ bW1 À cðW þ W0Þ ¼ 0:Solving this equation for FM will yieldFM ¼cðW þ W0Þ À bW1a: (i)For given numerical values of, b; c; y; W0; W1, and W,Eq. (i) can be used to determine the magnitude of theresultant muscle force. Once FM is calculated, its componentsin the x and y directions can be determined usingFMx ¼ FM sin y: (ii)FMy ¼ FM cos y: (iii)The horizontal and vertical components of the reactionforce developed at the sacrum can now be determined byutilizing the translational equilibrium conditions of thelower body of the athlete in the x and y directions:XFx ¼ 0 That is; FMx À FJx ¼ 0; then FJx ¼ FMx: (iv)XFy ¼ 0 That is; FMy À FJy À W1 þ ðW þ W0Þ ¼ 0;then FJy ¼ FMy þ W þ W0 À W1:(v)Assume that at an instant the athlete is bent so that histrunk makes an angle y ¼ 45with the vertical, and that thelengths of the lever arms are measured in terms of the heighth of the athlete and the weights are given in terms of theweight W of the athlete as a ¼ 0:02h; b ¼ 0:08h;c ¼ 0:12h; W0 ¼ W; and W1 ¼ 0:4W: Using Eq. (i)FM ¼ð0:12hÞðW þ WÞ À ð0:08hÞð0:4WÞ0:02h¼ 10:4 W:Fig. 5.19 Forces acting on the lower body of the athleteFig. 5.20 Free-body diagram5.7 Mechanics of the Spinal Column 73
  14. 14. From Eqs. (ii) and (iii)FMx ¼ ð10:4 WÞ ðsin 45Þ ¼ 7:4 WFMy ¼ ð10:4 WÞ ðcos 45Þ ¼ 7:4 W:From Eqs. (iv) and (v)FJx ¼ 7:4 WFJy ¼ 7:4 W þ W þ W À 0:4 W ¼ 9:0 W:Therefore, the magnitude of the resultant force on thesacrum isFJ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðFJxÞ2þ ðFJyÞ2q¼ 11:7 W:A remark• The results obtained are quite significant. While the ath-lete is bent forward by 45and lifting a weight withmagnitude equal to his own body weight, the erectorspinae muscles exert a force more than ten times theweight of the athlete and the force applied to the unionof the sacrum and the fifth lumbar vertebra is about 12times that of the body weight.5.8 Mechanics of the HipThe articulation between the head of the femur and theacetabulum of the pelvis (Fig. 5.21) forms a diarthrodialjoint. The stability of the hip joint is provided by its rela-tively rigid ball-and-socket type of configuration, itsligaments, and by the large and strong muscles crossing it.The femoral head fits well into the deep socket of theacetabulum. The ligaments of the hip joint, as well asthe labrum (a flat rim of fibrocartilage), support and holdthe femoral head in the acetabulum as the femoral headmoves. The construction of the hip joint is such that it isvery stable and has a great deal of mobility, thereby allowinga wide range of motion required for activities such as walk-ing, sitting, and squatting. Movements of the femur about thehip joint include flexion and extension, abduction andadduction, and inward and outward rotation. In someinstances, the extent of these movements is constrained byligaments, muscles, and/or the bony structure of the hip.The articulating surfaces of the femoral head and the acetab-ulum are lined with hyaline cartilage. Derangements of thehip can produce altered force distributions in the joint carti-lage, leading to degenerative arthritis.The pelvis consists of the ilium, ischium, and pubisbones, and the sacrum. At birth and during growth, thebones of the pelvis are distinct. In adults, the bones of thepelvis are fused and form synarthrodial joints that allow nomovement. The pelvis is located between the spine and thetwo femurs. The position of the pelvis makes it relativelyless stable. Movements of the pelvis occur primarily for thepurpose of facilitating the movements of the spine or thefemurs. There are no muscles whose primary purpose is tomove the pelvis. Movements of the pelvis are caused by themuscles of the trunk and the hip.Based on their primary actions, the muscles of the hipjoint can be divided into several groups (Fig. 5.22). Thepsoas, iliacus, rectus femoris, pectineus, and tensor fascialatae are the primary hip flexors. They are also used to carryout activities such as running or kicking. The gluteusmaximus and the hamstring muscles (the biceps femoris,semitendinosus, and semimembranosus) are hip extensors.The hamstring muscles also function as knee flexors. Thegluteus medius and gluteus minimus are hip abductormuscles providing for the inward rotation of the femur.The gluteus medius is also the primary muscle groupstabilizing the pelvis in the frontal plane. The adductorlongus, adductor brevis, adductor magnus, and gracilismuscles are the hip adductors. There are also small, deeplyplaced muscles (outward rotators) that provide for the out-ward rotation of the femur.The hip muscles predominantly suffer contusions andstrains occurring in the pelvis region.Example 5.5 During walking and running, we momentarilyput all of our body weight on one leg (the right leg inFig. 5.23). The forces acting on the leg carrying the totalbody weight are shown in Fig. 5.24 during such a single-legstance. FM is the magnitude of the resultant force exerted bythe hip abductor muscles, FJ is the magnitude of the jointreaction force applied by the pelvis on the femur, W1 is theweight of the leg, W is the total weight of the body applied asa normal force by the ground on the leg. The angle betweenthe line of action of the resultant muscle force and thehorizontal is designated by y.Fig. 5.21 Pelvis and the hip: (1) ilium, (2) sacrum, (3) acetabulum,(4) ischium, (5) greater trochanter, (6) lesser trochanter, (7) femur74 5 Applications of Statics to Biomechanics
  15. 15. A mechanical model of the leg, rectangular componentsof the forces acting on it, and the parameters necessary todefine the geometry of the problem are shown in Fig. 5.25.O is a point along the instantaneous axis of rotation of the hipjoint, point A is where the hip abductor muscles are attachedto the femur, point B is the center of gravity of the leg, andpoint C is where the ground reaction force is applied on thefoot. The distances between point A and points O, B, and Care specified as a, b, and c, respectively. a is the angle ofinclination of the femoral neck to the horizontal, and b is theangle that the long axis of the femoral shaft makes with thehorizontal. Therefore, a þ b is approximately equal to thetotal neck-to-shaft angle of the femur.Determine the force exerted by the hip abductor musclesand the joint reaction force at the hip to support the leg andthe hip in the position shown.Solution 1: Utilizing the Free-Body Diagram of the LegFor the solution of the problem, we can utilize the free-bodydiagram of the right leg supporting the entire weight of theperson. In Fig. 5.25a, the muscle and joint reaction forces areshown in terms of their components in the x and y directions.The resultant muscle force has a line of action that makes anangle y with the horizontal. Therefore,FMx ¼ FM cos y: (i)Fig. 5.24 Forces acting on the right leg carrying the entire weight ofthe bodyFig. 5.22 Muscles of the hip (a) anterior and (b) posterior views: (1)psoas, (2) iliacus, (3) tensor fascia latae, (4) rectus femoris, (5) sarto-rius, (6) gracilis, (7) gluteus minimis, (8) pectineus, (9) adductors, (10),(11) gluteus maximus medius, (12) lateral rotators, (13) bicepsfemoris, (14) semitendinosus, (15) semimembranosusFig. 5.23 Single-leg stance5.8 Mechanics of the Hip 75
  16. 16. FMy ¼ FM sin y: (ii)Since angle y is specified (given as a measured quantity),the only unknown for the muscle force is its magnitude FM.For the joint reaction force, neither the magnitude nor thedirection is known. With respect to the axis of the hip jointlocated at point O, ax in Fig. 5.25b is the moment arm of thevertical component FMy of the muscle force, and ay is themoment arm of the horizontal component of the muscleforce FMx. Similarly, ðbx À axÞ is the moment arm for W1and ðcx À axÞ is the moment arm for the force W applied bythe ground on the leg.From the geometry of the problemax ¼ a cos a: (iii)ay ¼ a sin a: (iv)bx ¼ b cos b: (v)cx ¼ c cos b: (vi)Now that the horizontal and vertical components of allforces involved, and their moment arms with respect to pointO are established, the condition for the rotational equilib-rium of the leg about point O can be utilized to determine themagnitude of the resultant muscle force applied at point A.Assuming that the clockwise moments are positive:XMo ¼ 0 : axFMy À ayFMx À ðcx À axÞWþ ðbx À axÞW1 ¼ 0:Substituting Eqs. (i) through (vi) into the above equation:ða cos aÞðFM sin yÞ À ða sin aÞðFM cos yÞÀ ðc cos b À a cos aÞW þ ðb cos b À a cos aÞ W1 ¼ 0:Solving this equation for the muscle force:FM ¼ðc W À bW1Þ cos b À aðW À W1Þ cos aaðcos a sin y À sin a cos yÞ: (vii)Notice that the denominator of Eq. (vii) can be simplifiedas a sinðy À aÞ. To determine the components of the jointreaction force, we can utilize the horizontal and verticalequilibrium conditions of the leg:XFx ¼ 0 : FJx ¼ FMx ¼ FM cos y: (viii)XFy ¼ 0 : FJy ¼ FMy þ W À W1FJy ¼ FM sin y þ W À W1: (ix)Therefore, the resultant force acting at the hip joint isFJ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðFJxÞ2þ ðFJyÞ2q: (x)Assume that the geometric parameters of the problemand the weight of the leg are measured in terms of theperson’s height h and total weight W as follows:a ¼ 0:05h; b ¼ 0:20h; c ¼ 0:52h; a ¼ 45; b ¼ 80; y ¼ 70;and W1 ¼ 0:17W. The solution of the above equations forthe muscle and joint reaction forces will yield FM ¼ 2:6Wand FJ ¼ 3:4W, the joint reaction force making an angle’ ¼ tanÀ1ðFJy=FJxÞ ¼ 74:8with the horizontal.Solution 2: Utilizing the Free-Body Diagram of theUpper Body Here, we have an alternative approach to thesolution of the same problem. In this case, instead of the free-body diagram of the right leg, the free-body diagram of theupper body (including the left leg) is utilized. The forcesacting on the upper body are shown in Figs. 5.26 and 5.27.FM is the magnitude of the resultant force exerted by the hipabductor muscles applied on the pelvis at point D. y is againthe angle between the line of action of the resultant muscleforce and the horizontal. FJ is the magnitude of the reactionforce applied by the head of the femur on the hip joint atFig. 5.25 Free-body diagram of the leg (a) and the geometricparameters (b)76 5 Applications of Statics to Biomechanics
  17. 17. point E. W2 ¼ W À W1 (total body weight minus the weightof the right leg) is the weight of the upper body and the left legacting as a concentrated force at point G. Note that point G isnot the center of gravity of the entire body. Since the right legis not included in the free-body, the left-hand side of the bodyis “heavier” than the right-hand side, and point G is located tothe left of the original center of gravity (a point along thevertical dashed line in Fig. 5.27) of the person. The locationof point G can be determined utilizing the method provided inSect. 4.12.By combining the individual weights of the segmentsconstituting the body under consideration, the problem isreduced to a three-force system. It is clear from the geometryof the problem that the forces involved do not form a parallelsystem. Therefore, for the equilibrium of the body, they haveto form a concurrent system of forces. This implies that thelines of action of the forces must have a common point ofintersection (point Q in Fig. 5.27), which can be obtained byextending the lines of action of W2 and FM. A line passingthrough points Q and E designates the line of action of thejoint reaction force Fj. The angle ’ that Fj makes withthe horizontal can now be measured from the geometryof the problem. Since the direction of Fj is determinedthrough certain geometric considerations, the number ofunknowns is reduced by one. As illustrated in Fig. 5.28,the unknown magnitudes FM and FJ of the muscle andjoint reaction forces can now be determined simply bytranslating W2, FM, and Fj to point Q, and decomposingthem into their components along the horizontal (x) andvertical (y) directions:FMx ¼ FM cos y;FMy ¼ FM sin y;FJx ¼ FJ cos ’;FJy ¼ FJ sin ’:(xi)For the translational equilibrium in the x and y directionsXFx ¼ 0 That is, À FMx þ FJx ¼ 0;then FJx ¼ FMx:Fig. 5.26 Forces acting on the pelvis during a single-leg (right leg)stanceFig. 5.27 Forces involved form a concurrent systemFig. 5.28 Resolution of the forces into their components5.8 Mechanics of the Hip 77
  18. 18. XFy ¼ 0 That is, FJy À W2 À FMy ¼ 0;then FJy ¼ FMy þ W2:Considering Eq. (xi)FJ cos ’ ¼ FM cos y; and (xii)FJ sin ’ ¼ FM sin y þ W2: (xiii)From Eq. (xii)FJ ¼FM cos ycos ’:Substituting this equation into Eq. (xiii) will yieldFM cos ycos ’sin ’ ¼ FM sin y þ W2;that isFM cos ycos ’sin ’ À FM sin y ¼ W2;FMcos y sin ’ À sin y cos ’cos ’ ¼ W2;thenFM ¼cos ’ W2sin ð’ À yÞ;andFJ ¼cos y W2sin ð’ À yÞ:For example, if y ¼ 70, ’ ¼ 74:8, and W2 ¼ 0:83 W(W is the total weight of the person), then the last twoequations will yield FM ¼ 2:6 W and FJ ¼ 3:4 W.How would the muscle and hip joint reaction forces varyif the person is carrying a load of W0 in each hand duringsingle-leg stance (Fig. 5.29)?The free-body diagram of the upper body while the per-son is carrying a load of W0 in each hand is shown inFig. 5.30. The system to be analyzed consists of the upperbody of the person (including the left leg) and the loadscarried in each hand. To counter-balance both the rotationaland translational (downward) effects of the extra loads, thehip abductor muscles will exert additional forces, and therewill be larger compressive forces generated at the hip joint.In this case, the number of forces is five. The gravitationalpull on the upper body (W2) and on the masses carried in thehands ðW0Þ form a parallel force system. If these parallelforces can be replaced by a single resultant force, then thenumber of forces can be reduced to three, and the problemcan be solved by applying the same technique explainedabove (Solution 2). For this purpose, consider the forcesystem shown in Fig. 5.31. Points M and N correspond tothe right and left hands of the person where external forcesof equal magnitude ðW0Þ are applied. Point G is the center ofgravity of the upper body including the left leg. The verticalFig. 5.29 Carrying a load in each handFig. 5.30 Forces acting on the upper bodyFig. 5.31 W3 is the resultant of the three-force system78 5 Applications of Statics to Biomechanics
  19. 19. dashed line shows the symmetry axis (midline) of the personin the frontal plane, and point G is located to the left of thisaxis. Note that the distance l1 between points M and G isgreater than the distance l2 between points N and G. If l1, l2,W2, and W0 are given, then a new center of gravity (point G0)can be determined by applying the technique of finding thecenter of gravity of a system composed of a number of partswhose centers of gravity are known (see Sect. 4.12). Byintuition, point G0is located somewhere between the sym-metry axis and point G. In other words, G0is closer to theright hip joint, and therefore, the length of the moment armof the total weight as measured from the right hip joint isshorter as compared to the case when there is no load carriedin the hands. On the other hand, the magnitude of theresultant gravitational force is W3 ¼ W2 þ 2W0, whichovercompensates for the advantage gained by the reductionof the moment arm.Once the new center of gravity of the upper body isdetermined, including the left leg and the loads carried ineach hand, Eqs. (xi) and (xii) can be utilized to calculate theresultant force exerted by the hip abductor muscles and thereaction force generated at the hip joint:FM ¼cos ’0ðW2 þ 2W0Þcos y sin ’0 À sin y cos ’0:Fj ¼cos y ðW2 þ 2W0Þcos y sin ’0 À sin y cos ’0:Here, Eqs. (xi) and (xii) are modified by replacing theweight W2 of the upper body with the new total weightW3 ¼ W2 þ 2W0, and by replacing the angle ’ that the lineof action of the joint reaction force makes with the horizon-tal with the new angle ’0(Fig. 5.32). ’0is slightly larger than’ because of the shift of the center of gravity from point G topoint G0toward the right of the person. Also, it is assumedthat the angle y between the line of action of the muscleforce and the horizontal remains unchanged.What happens if the person is carrying a load of W0 in theleft hand during a right-leg stance (Fig. 5.33)?Assuming that the system we are analyzing consists of theupper body, left leg, and the load in hand, the extra load W0carried in the left hand will shift the center of gravity ofthe system from point G to point G00toward the left of theperson. Consequently, the length of the lever arm of thetotal gravitational force W4 ¼ W2 þ W0 as measured fromthe right hip joint (Fig. 5.34) will increase. This will requirelarger hip abductor muscle forces to counter-balance theclockwise rotational effect of W4 and also increase the com-pressive forces at the right hip joint.It can be observed from the geometry of the systemanalyzed that a shift in the center of gravity from point Gto point G00toward the left of the person will decrease theangle between the line of action of the joint reaction forceand the horizontal from ’ to ’00. For the new configuration ofthe free-body shown in Fig. 5.34, Eqs. (xi) and (xii) can againbe utilized to calculate the required hip abductor muscleFig. 5.32 The problem is reduced to a three-force concurrent systemFig. 5.33 Carrying a load in one handFig. 5.34 Forces acting on the upper body5.8 Mechanics of the Hip 79
  20. 20. force and joint reaction force produced at the right hip(opposite to the side where the load is carried):FM ¼cos ’00ðW2 þ W0Þcos y sin ’00 À sin y cos ’00:FJ ¼cos y ðW2 þ W0Þcos y sin ’00 À sin y cos ’00:Remarks• When the body weight is supported equally on both feet,half of the supra-femoral weight falls on each hip joint.During walking and running, the entire mass of the bodyis momentarily supported by one joint, and we haveanalyzed some of these cases.• The above analyses indicate that the supporting forcesrequired at the hip joint are greater when a load is carriedon the opposite side of the body as compared to the forcesrequired to carry the load when it is distributed on eitherside. Carrying loads by using both hands and by bringingthe loads closer to the midline of the body is effective inreducing required musculoskeletal forces.• While carrying a load on one side, people tend to leantoward the other side. This brings the center of gravity ofthe upper body and the load being carried in the handcloser to the midline of the body, thereby reducing thelength of the moment arm of the resultant gravitationalforce as measured from the hip joint distal to the load.• People with weak hip abductor muscles and/or painfulhip joints usually lean toward the weaker side and walkwith a so-called abductor gait. Leaning the trunk side-ways toward the affected hip shifts the center of gravityof the body closer to that hip joint, and consequentlyreduces the rotational action of the moment of the bodyweight about the hip joint by reducing its moment arm.This in return reduces the magnitude of the forcesexerted by the hip abductor muscles required to stabilizethe pelvis.• Abductor gait can be corrected more effectively with acane held in the hand opposite to the weak hip, as com-pared to the cane held in the hand on the same side as theweak hip.5.9 Mechanics of the KneeThe knee is the largest joint in the body. It is a modifiedhinge joint. In addition to flexion and extension action of theleg in the sagittal plane, the knee joint permits some auto-matic inward and outward rotation. The knee joint isdesigned to sustain large loads. It is an essential componentof the linkage system responsible for human locomotion.The knee is extremely vulnerable to injuries.The knee is a two-joint structure composed of thetibiofemoral joint and the patellofemoral joint (Fig. 5.35).The tibiofemoral joint has two distinct articulations betweenthe medial and lateral condyles of the femur and the tibia.These articulations are separated by layers of cartilage,called menisci. The lateral and medial menisci eliminatebone-to-bone contact between the femur and the tibia, andfunction as shock absorbers. The patellofemoral joint is thearticulation between the patella and the anterior end of thefemoral condyles. The patella is a “floating” bone kept inposition by the quadriceps tendon and the patellar ligament.It increases the mechanical advantage of the quadricepsmuscle, improving its pulling effect on the tibia via thepatellar tendon. The stability of the knee is provided by anintricate ligamentous structure, the menisci and the musclescrossing the joint. Most knee injuries are characterized byligament and cartilage damage occurring on the medial side.The muscles crossing the knee protect it, provide internalforces for movement, and/or control its movement. Themuscular control of the knee is produced primarily by thequadriceps muscles and the hamstring muscle group(Fig. 5.36). The quadriceps muscle group is composed ofthe rectus femoris, vastus lateralis, vastus medialis, andvastus intermedius muscles. The rectus femoris muscle hasattachments at the anterior–inferior iliac spine and thepatella, and its primary actions are the flexion of the hipand the extension of the knee. The vastus lateralis, medialis,and intermedius muscles connect the femur and tibia throughthe patella, and they are all knee extensors. The bicepsfemoris, semitendinosus, and semimembranosus musclesmake up the hamstring muscle group, which help controlthe extension of the hip, flexion of the knee, and someinward–outward rotation of the tibia. Semitendinosus andsemimembranosus muscles have proximal attachmentson the pelvic bone and distal attachments on the tibia.Fig. 5.35 The knee: (1) femur, (2), medial condyle, (3) lateral con-dyle, (4) medial meniscus, (5) lateral meniscus, (6) tibial collateralligament, (7) fibular collateral ligament, (8) tibia, (9) fibula, (10)quadriceps tendon, (11) patella, (12) patellar ligament80 5 Applications of Statics to Biomechanics
  21. 21. The biceps femoris has proximal attachments on the pelvicbone and the femur, and distal attachments on the fibula.There is also the popliteus muscle that has attachments onthe femur and tibia. The primary function of this muscle isknee flexion. The other muscles of the knee are sartorius,gracilis, gastrocnemius, and plantaris.Example 5.6 Consider a person wearing a weight boot, andfrom a sitting position, doing lower leg flexion/extensionexercises to strengthen the quadriceps muscles (Fig. 5.37).Forces acting on the lower leg and a simple mechanicalmodel of the leg are illustrated in Fig. 5.38. W1 is the weightof the lower leg, W0 is the weight of the boot, FM is themagnitude of the tensile force exerted by the quadricepsmuscle on the tibia through the patellar tendon, and FJ isthe magnitude of the tibiofemoral joint reaction forceapplied by the femur on the tibial plateau. The tibiofemoraljoint center is located at point O, the patellar tendon isattached to the tibia at point A, the center of gravity of thelower leg is located at point B, and the center of gravity ofthe weight boot is located at point C. The distances betweenpoint O and points A, B, and C are measured as a, b, and c,respectively. For the position of the lower leg shown, thelong axis of the tibia makes an angle b with the horizontal,and the line of action of the quadriceps muscle force makesan angle y with the long axis of the tibia.Assuming that points O, A, B, and C all lie along astraight line, determine FM and FJ in terms ofa; b; c; y; b; W1, and W0.Solution Horizontal (x) and vertical (y) components of theforces acting on the leg and their lever arms as measuredfrom the knee joint located at point O are shown in Fig. 5.39.The components of the muscle force areFMx ¼ FM cosðy þ bÞ (i)FMy ¼ FM sinðy þ bÞ (ii)Fig. 5.37 Exercising the muscles around the knee jointFig. 5.38 Forces acting on the lower legFig. 5.36 Muscles of the knee: (1) rectus femoris, (2) vastus medialis,(3) vastus intermedius, (4) vastus lateralis, (5) patellar ligament, (6)semitendinosus, (7) semimembranosus, (8) biceps femoris, (9)gastrocnemius5.9 Mechanics of the Knee 81
  22. 22. There are three unknowns, namely FM, FJx, and FJy. Forthe solution of this two-dimensional (plane) problem, allthree equilibrium conditions must be utilized. Assumingthat the counter-clockwise moments are positive, considerthe rotational equilibrium of the lower leg about point O:XM0 ¼ 0 : ða cos bÞFMy À ða sin bÞFMxÀ ðb cos bÞW1 À ðc cos bÞ W0 ¼ 0Substituting Eqs. (i) and (ii) into the above equation, andsolving it for FM will yieldFM ¼ðbW1 þ cW0Þ cos ba cos b sinðy þ bÞ À sin b cosðy þ bÞ½ Š(iii)Note that this equation can be simplified by consideringthat cos b sinðy þ bÞ À sin b cosðy þ bÞ½ Š ¼ sin y; that isFM ¼bW1 þ cW0 cos ba sin yEquation (iii) yields the magnitude of the force that mustbe exerted by the quadriceps muscles to support the leg whenit is extended forward making an angle b with the horizontal.Once FM is determined, the components of the reaction forcedeveloped at the knee joint along the horizontal and verticaldirections can also be evaluated by considering the transla-tional equilibrium of the lower leg in the x and y directions:XFx ¼ 0 : FJx ¼ FMx ¼ FM cosðy þ bÞXFy ¼ 0 : FJy ¼ FMy À W0 À W1FJy ¼ FM sinðy þ bÞ À W0 À W1The magnitude of the resultant compressive force appliedon the tibial plateau at the knee joint isFJ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðFjxÞ2þ ðFjyÞ2q(iv)Furthermore, once Fjx and Fjy are determined, an angle’ that the joint reaction force FJ makes with the horizontalcan also be determined:’ ¼ arc tanFJyFJx Assume that the geometric parameters and the weightsinvolved are given as a ¼ 12 cm; b ¼ 22 cm; c ¼ 50 cm;W1 ¼ 150 N; W0 ¼ 100 N; y ¼ 15; and b ¼ 45, then byusing Eqs. (iii) and (iv):FM ¼ 1; 893 N; FJ ¼ 1; 681 N; ’ ¼ 55:7Remarks• The force FM exerted by the quadriceps muscle on thetibia through the patellar tendon can be expressed interms of two components normal and tangential to thelong axis of the tibia (Fig. 5.40). The primary function ofthe normal component FMn of the muscle force is to rotatethe tibia about the knee joint, while its tangential compo-nent FMt tends to translate the lower leg in a directioncollinear with the long axis of the tibia and applies acompressive force on the articulating surfaces of thetibiofemoral joint. Since the normal component of FM isa sine function of angle y, a larger angle between thepatellar tendon and the long axis of the tibia indicates alarger rotational effect of the muscle exertion. Thisimplies that for large y, less muscle force is wasted tocompress the knee joint, and a larger portion of themuscle tension is utilized to rotate the lower leg aboutthe knee joint.• One of the most important biomechanical functions of thepatella is to provide anterior displacement of the quadri-ceps and patellar tendons, thus lengthening the lever armof the knee extensor muscle forces with respect to theFig. 5.39 Force components, and their lever armsFig. 5.40 Rotational and translatory components of FM82 5 Applications of Statics to Biomechanics
  23. 23. center of rotation of the knee by increasing angle y(Fig. 5.41a). Surgical removal of the patella brings thepatellar tendon closer to the center of rotation of the kneejoint (Fig. 5.41b), which causes the length of the leverarm of the muscle force to decrease ðd2d1Þ. Losing theadvantage of having a relatively long lever arm, thequadriceps muscle has to exert more force than normalto rotate the lower leg about the knee joint.• The human knee has a two-joint structure composed of thetibiofemoral and patellofemoral joints. Notice that the quad-riceps muscle goes over the patella, and the patella and themuscle form a pulley–rope arrangement. The higher thetension in the muscle, the larger the compressive force(pressure) the patella exerts on the patellofemoral joint.We have analyzed the forces involved around thetibiofemoral joint by considering the free-body diagram ofthe lower leg. Having determined the tension in the patellartendon, and assuming that the tension is uniform throughoutthe quadriceps, we can calculate the compressive forceapplied on the patellofemoral joint by considering the free-body diagram of the patella (Fig. 5.42). Let FM be theuniform magnitude of the tensile force in the patellar andquadriceps tendons, Fp be the magnitude of the force exertedon the patellofemoral joint, a be the angle between thepatellar tendon and the horizontal, g be the angle betweenthe quadriceps tendon and the horizontal, and ’ be theunknown angle between the line of action of the compres-sive reaction force at the joint (Fig. 5.42b) and the horizon-tal. We have a three-force system and for the equilibrium ofthe patella it has to be concurrent.We can first determine the common point of intersectionQ by extending the lines of action of patellar and quadricepstendon forces. A line connecting point Q and the point ofapplication of Fp will correspond to the line of action of Fp.The forces can then be translated to point Q (Fig. 5.42c), andthe equilibrium equations can be applied. For the equilib-rium of the patella in the x and y directions:XFx ¼ 0 : Fp cos ’ ¼ FM ðcos g À cos aÞ (v)XFy ¼ 0 : Fp sin ’ ¼ FM ðsin a À sin gÞ (vi)These equations can be solved simultaneously for angle ’and the magnitude Fp of the compressive force applied bythe femur on the patella at the patellofemoral joint:From Eq. (v)Fp ¼FMðcos g À cos gÞcos ’;From Eq. (vi)Fp ¼FMðsin g À sin gÞsin ’; that isFig. 5.41 Patella increases the length of the lever armFig. 5.42 Static analysis of the forces acting on the patella5.9 Mechanics of the Knee 83
  24. 24. FMðcos g À cos gÞcos ’¼FMðsin g À sin gÞsin ’; then:sin ’ðcos g À cos gÞ ¼ cos ’ðsin g À sin gÞ;tan ’ ¼sin g À sin gcos g À cos gand’ ¼ tanÀ1 sin g À sin gcos g À cos g Once angle ’ is determined, then the magnitude of forceexerted on the patellofemoral joint Fp can also bedetermined:Fp ¼ FMcos g À cos gcos ’ :5.10 Mechanics of the AnkleThe ankle is the union of three bones: the tibia, fibula, andthe talus of the foot (Fig. 5.43). Like other major joints in thelower extremity, the ankle is responsible for load-bearingand kinematic functions. The ankle joint is inherently morestable than the knee joint which requires ligamentous andmuscular restraints for its stability.The ankle joint complex consists of the tibiotalar,fibulotalar, and distal tibiofibular articulations. The ankle(tibiotalar) joint is a hinge or ginglymus-type articulationbetween the spool-like convex surface of the trochlea of thetalus and the concave distal end of the tibia. Being a hingejoint, the ankle permits only flexion–extension (dorsiflex-ion–plantar flexion) movement of the foot in the sagittalplane. Other foot movements include inversion and eversion,inward and outward rotation, and pronation and supination.These movements occur about the foot joints such as thesubtalar joint between the talus and calcaneus and the trans-verse tarsal joints, talonavicular and calcaneocuboid.The ankle mortise is maintained by the shape of the threearticulations, and the ligaments and muscles crossing thejoint. The integrity of the ankle joint is improved by themedial (deltoid) and lateral collateral ligament systems, andthe interosseous ligaments. There are numerous musclegroups crossing the ankle. The most important ankle plantarflexors are the gastrocnemius and soleus muscles (Fig. 5.44).Both the gastrocnemius and soleus muscles are located in theposterior compartment of the leg and have attachments to theposterior surface of the calcaneus via the Achilles tendon.The gastrocnemius crosses the knee and ankle joints and hasfunctions in both. In the knee, it collaborates with kneeflexion and in the ankle it is the main plantar flexor.The plantar extensors or dorsiflexors are anterior muscles.They are the tibialis anterior, extensor digitorum longus,extensor hallucis longus and peroneous tertius muscles. Theprimary function of the lateral muscles (the peroneus longusand peroneus brevis) is to exert and plantarflex the ankle.Fig. 5.43 The ankle and foot: (1) tibia, (2) fibula, (3) medial malleolus,(4) lateral malleolus, (5) talus, (6) calcaneusFig. 5.44 Ankle muscles, (a) posterior, (b) anterior and (c) lateralviews: (1) gastrocnemius, (2) soleus, (3) Achilles tendon, (4) tibialisanterior, (5) extensor digitorum longus, (6) extensor hallucis longus, (7)peroneus longus, (8) peroneus brevis84 5 Applications of Statics to Biomechanics
  25. 25. The ankle joint responds poorly to small changes in itsanatomical configuration. Loss of kinematic and structuralrestraints due to severe sprains can seriously affect ankle stabil-ity and can produce malalignment of the ankle joint surfaces.The most common ankle injury, inversion sprain, occurs whenthe body weight is forcefully transmitted to the ankle while thefoot is inverted (the sole of the foot facing inward).Example 5.7 Consider a person standing on tiptoe on one foot(a strenuous position illustrated). The forces acting on the footduring this instant are shown in Fig. 5.45. W is the person’sweight applied on the foot as the ground reaction force, FM isthe magnitude ofthe tensile force exerted bythe gastrocnemiusand soleus muscles on the calcaneus through the Achillestendon, and FJ is the magnitude of the ankle joint reactionforce applied by the tibia on the dome of the talus. The weightof the foot is small compared to the weight of the body and istherefore ignored. The Achilles tendon is attached to the calca-neus at point A, the ankle joint center is located at point B, andthe ground reaction force is applied on the foot at point C. Forthis position of the foot, it is estimated that the line of action ofthe tensile force in the Achilles tendon makes an angle y withthe horizontal, and the line of action of the ankle joint reactionforce makes an angle b with the horizontal.Assuming that the relative positions of points A, B, and Care known, determine expressions for the tension in theAchilles tendon and the magnitude of the reaction force atthe ankle joint.SolutionWe have a three-force system composed of muscle force FM,joint reaction force FJ, and the ground reaction force W.From the geometry of the problem, it is obvious that for theposition of the foot shown, the forces acting on the foot donot form a parallel force system. Therefore, the force systemmust be a concurrent one. The common point of intersection(point O in Fig. 5.45) of these forces can be determined byextending the lines of action of W and FM. A straight linepassing through both points O and B represents the line ofaction of the joint reaction force. Assuming that the relativepositions of points A, B, and C are known (as stated in theproblem), the angle (say b) of the line of action of the jointreaction force can be measured.Once the line of action of the joint reaction force isdetermined by graphical means, the magnitudes of the jointreaction and muscle forces can be calculated by translatingall three forces involved to the common point of intersectionat O (Fig. 5.46). The two unknowns FM and FJ can now bedetermined by applying the translational equilibriumconditions in the horizontal (x) and vertical (y) directions.For this purpose, the joint reaction and muscle forces mustbe decomposed into their rectangular components first:FMx ¼ FM cos yFMy ¼ FM sin yFJx ¼ FJ cos bFJy ¼ FJ sin b:Fig. 5.45 Forces acting on the foot form a concurrent system of forcesFig. 5.46 Components of the forces acting on the foot5.10 Mechanics of the Ankle 85
  26. 26. For the translational equilibrium of the foot in the hori-zontal and vertical directions:XFx ¼ 0 : FJx ¼ FMx; that is, FJ cos b ¼ FM cos yXFy ¼ 0 : FJy ¼ FMy þW; that is, FJ sinb ¼ FM sinyþWSimultaneous solutions of these equations will yield:FM ¼W cos bcos y sin b À sin y cos b;that is, FM ¼W cos bsinðb À yÞFJ ¼W cos ycos y sin b À sin y cos b;that is, Fj ¼W cos ysinðb À yÞFor example, assume that y ¼ 45and b ¼ 60. ThenFM ¼ 1:93 W FJ ¼ 2:73 W:86 5 Applications of Statics to Biomechanics

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