In addition to the human body as a whole, each segment, such as the arm or trunk, also has a defined center of mass. In the lower extremity, for example, the major segments include the thigh, shank (lower leg), and foot. Figure 4-1 shows the center of mass of these segments for the lower extremities of a sprinter, indicated by black circles. The location of the center of mass within each segment remains fixed, approximately at its midpoint. In contrast, however, the location of the center of mass of the entire lower extremity changes with a change in spatial configuration of the segments (compare red circles in Figure 4-1). As shown for the left (flexed) lower extremity, the specific configuration of the segments can displace the center of mass of the lower limb outside the body. Additional information regarding the center of mass of body segments is discussed later in this chapter under the topic of anthropometry.

FIGURE 4-1. Lower extremities of a sprinter are illustrated, showing the centers of mass for the thigh, shank (lower leg), and foot segments as black circles. The center of mass for each lower extremity is shown as a red circle. The center of mass of the sprinter’s left lower extremity exists outside of the body. The axis of rotation of the right hip is indicated by the smaller green circle.

The mass moment of inertia of a body is a quantity that indicates its resistance to a change in angular velocity. Unlike inertia, its linear counterpart, the mass moment of inertia depends not only on the mass of the body, but, perhaps more important, on the distribution of its mass with respect to an axis of rotation.⁷ (Mass moment of inertia is often indicated by I and is expressed in units of kilograms-meters squared [kg-m²]). Because most human motion is angular rather than linear, the concept of mass moment of inertia is very relevant and important. Consider again the two positions of the lower extremities of the sprinter in Figure 4-1. Within each segment, the individual centers of mass of the thigh, shank, and foot are in the same location in both lower extremities; however, because of the different degrees of knee flexion, the distances of the centers of mass of the shank and foot segments have changed relative to the hip joint. As a consequence the mass moment of inertia of each entire limb changes; the right extended (and “longer”) lower extremity has a greater mass moment of inertia than the left. (Another way of conceptualizing the increase is to note that as the knee extends, the center of mass of the entire right lower extremity, depicted by the red circle, moves farther from the hip, thereby increasing its mass moment of inertia.) The ability to actively change an entire limb’s mass moment of inertia can profoundly affect the muscle forces and joint torques necessary for movement. For example, during the swing phase of running, the entire lower limb functionally shortens by the combined movements of knee flexion and ankle dorsiflexion (as in the left lower extremity in Figure 4-1). The lower limb’s reduced mass moment of inertia reduces the torque required by the hip muscles to accelerate and decelerate the limb during swing phase. This concept can be readily appreciated during the swing phase while running with the knees held nearly extended (increased I), or almost fully flexed (decreased I).

The concept of mass moment of inertia applies to both rehabilitation and recreational settings. Consider, for example, the design of a prosthesis for the person with a lower limb amputation. The use of lighter components in the foot prosthesis, for example, not only reduces the overall mass (and weight) of the prosthesis, but also results in a change in the distribution of the mass to a more proximal location in the leg. As a result, less resistance is imposed on the remaining limb during the swing phase of gait. The benefit of these lighter components is realized in terms of lessened energy requirements for the person with an amputation. Changing footwear can also make a difference. Consider changes in the mass moment of inertia and resultant required torques for gait when the person changes from wearing a lightweight tennis shoe to a wearing a heavy winter boot.

Athletes often attempt to control the mass moment of inertia of their entire body by altering the position of their individual body segments relative to the axis of rotation. This concept is well illustrated by divers who reduce their moment of inertia in order to successfully complete multiple somersaults while in the air (Figure 4-2, A). The athlete can assume an extreme “tuck” position by placing the head near the knees, holding the arms and legs tightly together, thereby bringing their segments’ centers of gravity closer to the axis of rotation. Based on the principle of “conservation of angular momentum,” reducing the body’s mass moment of inertia results in an increased angular velocity. Conversely, the athlete could slow the rotation by assuming a “pike” (see Figure 4-2, B) position and increasing the body’s moment of inertia, or assuming a “layout” position (see Figure 4-2, C), which maximizes the body’s mass moment of inertia and greatly slows the body’s angular velocity.

FIGURE 4-2. A diver illustrates an example of how the mass moment of inertia around a medial-lateral axis (red circle) can be altered through changes in position of the extremities. In position A, the diver decreases the mass moment of inertia, which increases the angular velocity of the spin. In positions B and C, the extremities are positioned progressively farther from the axis, and the angular velocity is progressively slower.

SPECIAL FOCUS 4-1 A Closer Mathematic Look at the Concept of Mass Moment of Inertia

Thus far in this chapter, the concept of mass moment of inertia (I) has been described primarily from a functional standpoint. It may be instructive, however, to consider this physical property from a more mathematic perspective. I is formally defined in the following equation, in which n indicates the number of particles in a body, m_i is the mass of each particle in the body, and r_i is the distance of each particle to the axis of rotation.

Mass Moment of Inertia

(Equation 4.2)

As a way to further explore Equation 4.2, it will be used to determine how the grip applied to a baseball bat dramatically affects its mass moment of inertia and therefore the difficulty in swinging the bat. The bat illustrated in Figure 4-3 is considered to consist of six point masses (m₁ to m₆), ranging from 0.1 to 0.225 kg, each located 0.135 m from another. During the swing the batter rotates the bat; for illustration purposes the axis of this rotation is positioned as Y₁ (red line). If the bat is not sized correctly for the batter, the batter will often “choke up” by shifting his or her grip farther down the bat; again, for illustration purposes, the axis is now positioned as Y₂ (blue line). The calculations shown in the box demonstrate how the distribution of the mass particles, relative to a given axis of rotation, dramatically affects the mass moment of inertia of the rotating bat. First consider Y₁ as the axis of rotation. The mass moment of inertia of the bat is determined using Equation 4.2 and substituting known values. Next, consider Y₂ as the axis of rotation. The important point here is that the mass particles are distributed differently when each axis is considered separately. As seen in the calculations, the mass moment of inertia when considering Y₂ as the axis is 58% of that if Y₁ is the considered axis. This means that the batter could achieve the same angular acceleration with 58% less torque. Or, for the same torque, the bat would accelerate 1.72 times as fast. This is a significant functional advantage gained by choking up on the bat; the bat is easier to swing, although its mass and weight have not changed. The reason for the reduced moment of inertia is that mass points m₂ through m₆ are closer to the Y₂ axis. This is very significant mathematically when one considers that the mass moment of inertia of each point is related to the square of the distance to the axis.

FIGURE 4-3. A baseball bat is shown with a potential to rotate around two separate axes of rotation (Y₁, Y₂). The set of calculations associated with each axis of rotation shows how the distribution of mass relative to the axis of rotation affects the mass moment of inertia. The bat is assumed to consist of six mass points (m₁ to m₆), ranging from 0.1 to 0.225 kg, located at equal distances from one another.

Y₁ Axis of Rotation

Y₂ Axis of Rotation

The mass moment of inertia of human body segments is more difficult to determine than for the baseball bat, although they are based on the same mathematic principle. Much of the difficulty stems from the fact that segments in the human body are made up of different tissues, such as bone, muscle, fat, and skin and are not of uniform density. Values for the mass moment of inertia for each body segment have been generated from cadaver studies, mathematic modeling, and various imaging techniques.3,8

NEWTON’S SECOND LAW: LAW OF ACCELERATION

Force (Torque)-Acceleration Relationship: Newton’s second law states that the linear acceleration of a body is directly proportional to the force causing it, takes place in the same direction in which the force acts, and is inversely proportional to the mass of the body. Newton’s second law generates an equation that relates force (F), mass (m), and acceleration (a) (Equation 4.1). Conceptually, Equation 4.1 defines a force-acceleration relationship. Considered a cause-and-effect relationship, the left side of the equation, force (F), can be regarded as a cause because it represents a pull or push exerted on a body; the right side, m × a, represents the effect of the pull or push. In this equation, ΣF designates the sum of, or net, forces acting on a body. If the sum of the forces acting on a body is zero, acceleration is also zero and the body is in linear equilibrium. As previously discussed, this case is described by Newton’s first law. If, however, the net force produces acceleration, the body will accelerate in the direction of the resultant force. In this case, the body is no longer in equilibrium.

Newton’s Second Law of Linear Motion Quantifying a Force

(Equation 4.1)

Force is measured in newtons, where 1 newton (N) = 1 kg-m/sec².

The rotary or angular counterpart to Newton’s second law states that a torque will cause an angular acceleration of a body around an axis of rotation. Furthermore, the angular acceleration of a body is directly proportional to the torque causing it, takes place in the same rotary direction in which the torque acts, and is inversely proportional to the mass moment of inertia of the body. (The italicized words denote the essential differences between the linear and angular counterparts of this law.) For the rotary condition, Newton’s second law generates an equation that relates the torque (T), mass moment of inertia (I), and angular acceleration (α) (Equation 4.3). (This chapter uses the term torque. The reader should be aware that this term is interchangeable with terms moment and moment of force.) In this equation, ΣT designates the sum of, or net, torques acting to rotate a body. Conceptually, Equation 4.3 defines a torque–angular acceleration relationship. Within the musculoskeletal system, the primary torque producers are muscles. A contracting biceps muscle, for example, produces a net internal flexion torque at the elbow. Neglecting external influences such as gravity, the angular acceleration of the rotating forearm is proportional to the internal torque (i.e., the product of the muscle force and its internal moment arm) but is inversely proportional to the mass moment of inertia of the forearm-and-hand segment. Given a constant internal torque, the forearm-and-hand segment with the smaller mass moment of inertia will achieve a greater angular acceleration than one with a larger mass moment of inertia. (A smaller mass moment of inertia can be achieved by moving a cuff weight from the wrist to the mid-forearm, for example.) Understand that this inertial resistance to the angular acceleration of the limb applies even in the absence of gravity. For example, consider the positions of the lower limb in Figure 4-1 but with the person on his or her side in a “gravity eliminated” position. Because of changes in the mass moment of inertia, less muscular effort will be required to flex the hip with the knee also flexed than with the knee extended.

Newton’s Second Law of Rotary Motion Quantifying a Torque

(Equation 4.3)

Torque is expressed in newton-meters, where 1 Nm = 1 kg-m² × radians/sec².

Impulse-Momentum Relationship: Additional relationships can be derived from Newton’s second law through the broadening and rearranging of Equations 4.1 and 4.3. One such relationship is the impulse-momentum relationship.

Acceleration is the rate of change of velocity (Δv/t). Substituting this expression for linear acceleration in Equation 4.1 results in Equation 4.4. Equation 4.4 can be further rearranged to Equation 4.5.

(Equation 4.4)

(Equation 4.5)

Application of a linear impulse (force multiplied by time) leads to a change in linear momentum (mass multiplied by a change in linear velocity).

The product of mass and velocity on the right side of Equation 4.5 defines the momentum of a moving body. Momentum describes the quantity of motion possessed by a body. Momentum is generally represented by the letter p and has units of kilogram-meters per second (kg-m/sec). An impulse is a force applied over a period of time (the product of force and time on the left side of Equation 4.5). The linear momentum of an object such as a moving car is changed by the application of a force over a given time. When a quick change in momentum is required (during an emergency stop, for instance), a very large brake force is applied for a short time. Less brake force for the same time, or the same brake force for even less time, results in a smaller change in momentum. Impulse and momentum are vector quantities. Equation 4.5 defines the linear impulse-momentum relationship.

The impulse-momentum relationship provides another perspective from which to study human performance, as well as to gain insight into injury mechanisms. At certain locations the body develops mechanisms and structures to lessen peak external load forces. For example, when landing from a jump, peak forces can be reduced throughout the joints of the lower extremities if the impact of the landing is prolonged by more “give” in the muscles—through a lower level and prolonged eccentric activation. As another example, as the foot contacts the ground during normal gait, the fat pad over the plantar surface of the calcaneus cushions the interaction between the foot and the ground and works to decrease peak reaction forces. Running footwear often augments this function with shock-absorbing outsoles to further cushion the impact of the foot on the ground. Bicycle helmets, rubber or springed flooring, and protective padding are additional examples of equipment designs intended to reduce injuries by increasing the duration of impact in order to minimize the peak force of the impact.

Newton’s second law involving torque can also apply to the rotary case of the impulse-momentum relationship. Similar to the substitutions and rearrangements for the linear relationship, the angular relationship can be expressed by substitution and rearrangement of Equation 4.3. Substituting Δω/t (rate of change in angular velocity) for α (angular acceleration) results in Equation 4.6. Equation 4.6 can be rearranged to Equation 4.7—the angular equivalent of the impulse-momentum relationship. Torque and angular momentum are also vector quantities.

(Equation 4.6)

(Equation 4.7)

Application of an angular impulse (torque multiplied by time) leads to a change in angular momentum (mass moment of inertia times a change in angular velocity).

SPECIAL FOCUS 4-2 A Closer Look at the Impulse-Momentum Relationship

Numerically, an impulse can be calculated as the product of the average force (N) and its time of application. Impulse can also be represented graphically as the area under a force-time curve. Figure 4-4 displays a force-time curve of the horizontal component of the anterior-posterior shear force applied by the ground against the foot (ground reaction force) as an individual runs across a force plate embedded in the floor. The curve is biphasic: the posterior-directed impulse during initial floor contact is negative, and the anterior-directed impulse during propulsion is positive. If the two impulses (i.e., areas under the curves) are equal, the net impulse is zero, and there is no change in the momentum of the system. In this example, however, the posterior-directed impulse is greater than the anterior, indicating that the runner’s forward momentum is decreased.

FIGURE 4-4. Graphic representation of the areas under a force-time curve showing the (A) posterior-directed and (B) anterior-directed impulses of the horizontal component of the ground reaction force while running.

Work-Energy Relationship: To this point, Newton’s second law has been described using (1) the force (torque)-acceleration relationship (Equations 4.1 and 4.3) and (2) the impulse-momentum relationship (Equations 4.4 through 4.7). Newton’s second law can also be restated to provide a work-energy relationship. This third approach can be used to study human movement by analyzing the extent to which work causes a change in an object’s energy. Work occurs when a force or torque operates over some linear or angular displacement. Work (W) in a linear sense is equal to the product of the magnitude of the force (F) applied against an object and the linear displacement of the object in the direction of the applied force (Equation 4.8). If no movement occurs in the direction of the applied force, no mechanical work is done. Similar to the linear case, angular work can be defined as the product of the magnitude of the torque (T) applied against the object, and the angular displacement of the object in the direction of the applied torque (Equation 4.9). Work is expressed in joules (J).

SPECIAL FOCUS 4-3 Positive Work, Negative Work, and “Isometric Work”

As described, linear work is found by multiplying an applied force by the object’s displacement in the direction of the applied force. Consider, for example, the linear force applied by the contracting elbow flexor muscles to flex the elbow and bring the hand to the mouth. In a linear sense, the work is the product of the muscles’ contractile force and the distance the muscles shortened. In an angular sense, the rotary work is computed by the torque applied by the elbow flexors and the amount of flexion (in radians) occurring at the elbow. In this case the work is positive because the rotation of the forearm is in the same direction as the applied torque. In addition, the elbow flexors are shortening through concentric activation, and the work is being performed by the muscles. In contrast, when the elbow flexors are active, but the elbow is extending (e.g., when slowly lowering a weight), the elbow flexors are lengthening through an eccentric activation, and the work is being performed on the muscles. In this case, because the rotation is in a direction opposite to the applied torque, the work is negative. The final scenario is when the elbow flexors are active but no movement is taking place, as when a muscle is active isometrically. In this case, even though considerable metabolic energy may be expended, no mechanical work is being performed.

Relationships between Work (W), Potential Energy (PE), and Kinetic Energy (KE)

(Equation 4.8)

(Equation 4.9)

Just as the impulse-momentum relationship describes the change in momentum caused by a force applied over a given time, the work-energy relationship describes the change in kinetic energy caused by a force applied over a given displacement. Using the example described earlier can illustrate the similarity in these concepts. The kinetic energy of an object such as a moving car is changed by the application of a force over a displacement. When a quick change in kinetic energy is required (e.g., for an emergency stop), a very large brake force is applied over a short displacement. Less brake force for the same displacement or the same brake force applied for even less displacement results in a smaller change in kinetic energy. Work and displacement are vector quantities.

The work-energy relationship does not take into account the time over which the forces or torques are applied. Yet in most daily activities it is often the rate of performing work that is important. The rate of performing work is defined as power. The ability for muscles to generate adequate power may be critical to the success of movement or to the understanding of the impact of a treatment intervention. On the basketball court, for example, it is often the player’s vertical speed at takeoff that determines success in achieving a rebound. Another example of the importance of the rate of work can be appreciated in an elderly person with Parkinson’s disease who must cross a busy street in the time determined by a pedestrian traffic signal.

Average power (P) is work (W) divided by time (Equation 4.10). Because work is the product of force (F) and displacement (d), the rate of work at any instant can be restated in Equation 4.11 as the product of force and velocity. Angular power may also be defined as in the linear case, using the angular analogs of force and linear velocity: torque (T) and angular velocity (ω), respectively (Equation 4.12). Angular power is often used as a clinical measure of muscle performance. The mechanical power produced by the quadriceps, for example, is equal to the net internal torque produced by the muscle times the average angular velocity of knee extension. Power is often used to designate the net transfer of energy between active muscles and external loads. Positive power reflects the rate of work done by concentrically active muscles against an external load. Negative power, in contrast, reflects the rate of work done by the external load against eccentrically active muscles.

Power (P)

(Equation 4.10)

(Equation 4.11)

(Equation 4.12)

Table 4-2 summarizes the definitions and units needed to describe many of the physical measurements related to Newton’s second law.

TABLE 4-2.

Physical Measurements Associated with Newton’s Second Law

^*To convert from English units to SI units, multiply the English value by the number in the table cell. To convert from SI units to English, divide by that number. If two equations are in the cell, the upper equation is used to convert the linear measure, and the lower equation is used to convert the angular measure.

^†Radians, which are unitless, may be used instead of degrees (1 radian = about 57.3 degrees).

^‡The English unit of a mass is a pound-mass (lbm) or slug.

NEWTON’S THIRD LAW: LAW OF ACTION-REACTION

Newton’s third law of motion states that for every action there is an equal and opposite reaction. This law implies that every effect one body exerts on another is counteracted by an effect that the second body exerts on the first. The two bodies interact simultaneously, and the consequence is specified by Newton’s law of acceleration (ΣF = m × a); that is, each body experiences a different effect, and that effect depends on its mass. For example, a person who falls off the roof of a second-story building exerts a force on the ground, and the ground exerts an equal and opposite force on the person. Because of the huge discrepancies in mass between the earth and the person, the effect, or acceleration experienced by the person, is much greater than the effect “experienced” by the ground. As a result, the person may sustain significant injury.

Another example of Newton’s law of action-reaction is the reaction force provided by the surface on which one is walking or standing. The foot produces a force against the ground, and in accordance with Newton’s third law, the ground generates a ground reaction force in the opposite direction but of equal magnitude (Figure 4-5). The ground reaction force changes in magnitude, direction, and point of application on the foot throughout the stance period of gait. Newton’s third law also has an angular equivalent. For example, during an isometric exercise, the internal and external torques are equal and in opposite rotary directions.

FIGURE 4-5. The action of the forces produced between the ground and foot are illustrated during the contact phase of the “swing-through” method of crutch-assisted walking. The action of the foot forces (white arrows) is counteracted by the ground reaction forces (black arrows). If the horizontal component of the ground reaction force (caused by friction) is less than the horizontal component of the foot force, the foot will slide forward on the floor according to Newton’s second law: F = m × a.

INTRODUCTION TO MOVEMENT ANALYSIS: SETTING THE STAGE FOR ANALYSIS

The previous section describes the nature of the cause-and-effect relationships between force and motion as outlined by Newton’s laws. Now that this foundation has been established, this section introduces the steps and conventions used to formally analyze movement. Special attention is paid to the analysis of internal and external forces and torques and how these variables affect the body and its joints. This section should fully prepare the reader to follow the mathematic solutions to three sample problems constructed in the next section.

Anthropometry

Anthropometry is derived from the Greek root anthropos, man, + metron, measure. In the context of human movement analysis, anthropometry may be broadly defined as the measurement of certain physical design features of the human body, such as length, mass, weight, volume, density, center of gravity, and mass moment of inertia. Knowledge of these parameters is often essential to conducting kinematic and kinetic analyses for both normal and pathologic motion. Variables such as mass and mass moment of inertia of individual limb segments, for example, are needed to determine the inertial properties that muscles must overcome to generate movement. Anthropometric information is also valuable in the design of the work environment, furniture, tools, and sports equipment.

Much of the information pertaining to the body segments’ center of gravity and mass moment of inertia has been derived from cadaver studies.3,4 Other methods for deriving anthropometric data include mathematic modeling and imaging techniques, such as computed tomography and magnetic resonance imaging. Table 4-3 lists data on the weights of body segments and the location of the center of gravity. (The specific details contained in this table will be needed to solve selected parts of the biomechanical problems posed in Appendix I, Part B).

TABLE 4-3.

Anthropometric Data on Body Segments’ Weight and Center of Gravity Location in the Anatomic Position (Extremity Data Are Unilateral Only)

Compiled results in Winter DA: Biomechanics and motor control of human movement, ed 3, New York, 2005, John Wiley & Sons. Mass moments of inertia are not included in this table because the focus of this chapter is limited to static analysis.

^*Even though some definitions listed in this table do not represent the endpoints of the segment, they are easily identified locations on the human. The values for segment weight and center of gravity location in this table take into consideration the discrepancy between the definition of the segment and the true endpoints. For example, the segment definition for the forearm is the same for the forearm-and-hand, but the percentages listed for the segment weight and center of gravity location of the forearm-and-hand are higher, taking into consideration the mass of the hand.

Free Body Diagram

The analysis of movement requires that all forces that act on the body be taken into account. Before any analysis, a free body diagram is constructed to facilitate the process of solving biomechanical problems. The free body diagram is a “snapshot” or simplified sketch that represents the interaction between a body and its environment. The body under consideration may be a single rigid segment, such as the foot, or it may be several segments, such as the head, arms, and trunk. When the body consists of several segments, these are assumed to be rigidly connected together into a single rigid system.

A free body diagram requires that all relevant forces acting on the system are carefully drawn. These forces may be produced by muscle, gravity (as reflected in the weight of the segment), fluid, air resistance, friction, and ground reaction forces. Arrows are used to indicate force vectors.

How a free body diagram is configured depends on the intended purpose of the analysis. Consider the example presented in Figure 4-6. In this example, the free body diagram represents the shank-and-foot at the instant of initial heel contact during walking. The free body diagram involves figuratively “cutting through” the desired joint(s) to isolate or “free” the body of interest. In the example presented in Figure 4-6, the knee joint was cut through to isolate the shank-and-foot segment. The effects of active muscle force are usually distinguished from the effects of other soft tissues, such as passive tension created in stretched joint capsule and ligaments. Although the contribution of individual muscles acting across a joint may be determined, a single resultant muscle force (M) vector is often used to represent the sum total of all individual muscle forces. Other forces external to the system are added to the diagram, which may include the ground reaction force (G) and weight of the shank-and-foot segments (S and F). As specified by Newton’s third law, the ground reaction force is the force developed in response to the foot striking the earth.

FIGURE 4-6. Free body diagram of the shank-and-foot at the instant of heel contact during walking. The segment is isolated by figuratively “cutting through” the knee joint. Relevant forces are drawn in as shown. The X-Y coordinate reference frame is placed so the X axis is parallel with the shank.

An additional force is identified in Figure 4-6: the joint reaction force (J). This force includes joint contact forces as well as the net or cumulative effect of all other forces transmitted from one segment to another. Joint reaction forces are caused “in reaction” to other forces, such as those produced by activation of muscle, by passive tension in stretched periarticular connective tissues, and by gravity (body weight). As will be discussed, the free body diagram is completed by defining an X-Y coordinate reference frame and writing the governing equations of motion.

Clinically, reducing joint reaction force is often a major focus in treatment programs designed to lessen pain and prevent further joint degeneration in persons with arthritis. Frequently, treatments are directed toward reducing joint forces through changes in the magnitude of muscle activity and their activation patterns or through a reduction in the weight transmitted through a joint. Consider the patient with osteoarthritis of the hip joint as an example. The magnitude of the hip joint’s reaction force may be decreased by having the person reduce walking velocity, thereby lessening the magnitude of muscle activation. Highly cushioned shoes may be recommended to reduce impact forces. In addition, a cane may be used to reduce forces through the hip joint.1,10,13 If obesity is a factor, a weight-reduction program may be recommended.

STEPS FOR CONSTRUCTING THE FREE BODY DIAGRAM

The key elements needed to begin solving problems related to human movement are to determine the purpose of the analysis, identify the free body of interest, and indicate all the forces that act on that body. The following example presents steps to assist with construction of a free body diagram.

Consider the situation in which an individual is holding a weight out to the side, as shown in Figure 4-7. This free body is assumed to be in static equilibrium, and the sum of all forces and the sum of all torques acting on the body are equal to zero. One purpose of the analysis might be to determine how much muscle force is required by the glenohumeral joint abductor muscles (M) to keep the arm abducted to 90 degrees; another purpose might be to determine the magnitude of the glenohumeral joint reaction force (J) during this same activity.

FIGURE 4-7. A frontal plane depiction of a free body diagram isolating the system as a right arm and ball combination: resultant shoulder abductor muscle force (M); glenohumeral joint reaction force (J); segment weight (S); and ball weight (B). The axis of rotation is shown as an open red circle at the glenohumeral joint. The X-Y coordinate reference frame is placed so the X axis is parallel with the upper extremity. (Modified from LeVeau BF: Williams & Lissner’s biomechanics of human motion, ed 3, Philadelphia, 1992, Saunders.)

Step I of constructing the free body diagram is to identify and isolate the free body under consideration. In this example, the glenohumeral joint was “cut through,” and the free body is the combination of the entire arm and the resistance (exercise ball weight).

Step II involves defining a coordinate reference frame that allows the position and movement of a body to be defined with respect to a known point, location, or axis (see Figure 4-7, X-Y coordinate reference frame). More detail on establishing a reference frame is discussed ahead.

Step III involves identification and inclusion of all forces that act on the free body. Internal forces are those produced by muscle (M). External forces include the force of gravity on the mass of the exercise ball (B), as well as the force of gravity on the arm segment (S). Although not relevant to Figure 4-7, other examples of external forces could include forces applied by therapists, cables, resistance bands, the ground or other surface, air resistance, and orthotics. The forces are drawn on the figure while specifying their approximate point of application and spatial orientation. For example, vector S acts at the center of gravity of the upper extremity, a location determined by using anthropometric data, such as those presented in Table 4-3.

The direction of the muscle force (M) is drawn to correspond to the line of muscle pull and in a direction to generate torque that opposes the net torque produced by the external forces. In this example the torque produced by the external forces, S and B, tends to rotate the arm in a clockwise, adduction, or −Z direction. The line of force of M, therefore, in combination with its moment arm, creates a torque in a counterclockwise, abduction direction, or +Z direction. (The convention of using + or −Z to designate rotation direction is described ahead.)

Step IV of the procedure is to show the joint reaction force (J), in this case created across the glenohumeral articulation. Initially the direction of the joint reaction force may not be known, but, as explained later, it is typically drawn in a direction opposite to the pull of the dominant muscle force. The precise direction of J can be determined after static analysis is carried out and unknown variables are calculated.

Step V involves writing the three governing equations required to solve two-dimensional (2D) static equilibrium problems encountered in this chapter. The equations are: ΣTorque_Z = 0; ΣForce_X = 0; ΣForce_Y = 0. These equations are explained later in the chapter.

Steps in Constructing the Free Body Diagram

Step I: Identify and isolate the free body under consideration.

Step II: Establish a coordinate reference frame.

Step III: Draw the internal (muscular) and external forces that act on the system.

Step IV: Draw the joint reaction force.

Step V: Write the governing equations of motion.

SPATIAL REFERENCE FRAMES

In order to accurately describe motion or solve for unknown forces, a spatial reference frame needs to be established. This information allows the position and direction of movement of a body, a segment, or an object to be defined with respect to some known point, location, or segment’s axis of rotation. If a reference frame is not identified, it becomes very difficult to interpret and compare measurements in clinical and research settings.

A spatial reference frame is arbitrarily established and may be placed inside or outside the body. Reference frames used to describe position or motion may be considered either relative or global. A relative reference frame describes the position of one limb segment with respect to an adjacent segment, such as the foot relative to the leg, the forearm relative to the upper arm, or the trunk relative to the thigh. A measurement is made by comparing motion of an anatomic landmark or coordinates between segments of interest. Goniometry provides one example of a relative reference frame used in clinical practice. Elbow joint range of motion, for example, describes a measurement using a relative reference frame defined by the long axes of the upper arm and forearm segments, with an axis of rotation through the elbow.

Relative reference frames, however, lack the information needed to define motion with respect to a fixed point or location in space. To analyze motion with respect to the ground, direction of gravity, or another type of externally defined reference frame in space, a global (laboratory) reference frame must be defined. Excessive anterior or lateral deviations of the trunk during gait are examples of a measurement made with respect to a global reference frame. In these examples, the position of the trunk is measured with respect to an external vertical reference.

Whether motion is measured via a relative or global reference frame, the location of a point or segment in space can be specified using a coordinate reference frame. In laboratory-based human movement analysis, the Cartesian coordinate system is most frequently employed. The Cartesian system uses coordinates for locating a point on a plane in 2D space by identifying the distance of the point from each of two intersecting lines, or in three-dimensional (3D) space by the distance from each of three planes intersecting at a point. A 2D coordinate reference frame is defined by two imaginary axes arranged perpendicular to each other with the arrowheads pointed in positive directions. The two axes (labeled, for example, X and Y) may be oriented in any manner that facilitates quantitative solutions (compare Figures 4-6 and 4-7, for example). A 2D reference frame is frequently used when the motion being described is predominantly planar (i.e., in one plane), such as knee flexion and extension during gait.

In most cases, human motion occurs in more than one plane. In order to fully describe this type of motion, a 3D coordinate reference frame is necessary. A 3D reference frame typically has three axes (X, Y, and Z), each perpendicular (or orthogonal) to another. As with the 2D system, the arrowheads point in positive directions. A universal convention for orienting this triplanar coordinate system in space is based on the right-hand rule. This rule is used throughout most quantitative biomechanical studies (see Special Focus 4-4).

Throughout most of this textbook, the terminology used to describe linear direction within planes (such as the direction of a muscle force or an axis of rotation around a joint) is less formal than that dictated by the right-hand rule. As described in Chapter 1, linear direction in space is loosely described relative to the human body standing in the anatomic position, using terms such as anterior-posterior, medial-lateral, and vertical. Although useful for most qualitative or anatomic-based descriptions, this convention is not well suited for quantitative analyses, such as those introduced later in this chapter. In these cases, the Cartesian coordinate system is used, and the orientation of its 3D axes is designated by the right-hand rule.

SPECIAL FOCUS 4-4 The “Right-Hand Rule”: a Convention for Completely Describing the Spatial Orientation of a Three-Dimensional Coordinate Reference Frame

When a Cartesian coordinate system is set up, the direction or orientation of the orthogonal axes is not arbitrary. A convention must be used to facilitate the sharing of research from different laboratories throughout the worldwide scientific community. Using Figure 4-7 as an example, the X and Y axes are in the plane of the page or, relative to the subject, parallel with the frontal plane. (It is often most convenient, although not mandatory, to orient the X-Y axes so that the X axis is parallel with the body segment of interest.) A third axis, the Z axis, must be defined. Although not drawn in the figure, the Z axis is oriented perpendicular to the X-Y plane. By convention, the direction of the arrowheads shown on the X-Y coordinate reference frame indicates positive directions. As shown in Figure 4-7, positive X direction is to the right and positive Y direction is upward. The right-hand rule can be used to define the direction (+ or −) of the Z axis. Applying the right-hand rule is performed by laying the ulnar border of your right hand along the X axis, with the straight fingers pointing in a positive X direction (toward the ball on the model). Your hand should be positioned along the X axis so that when your fingers flex, they curl from the positive X direction toward the positive Y direction. Your extended thumb is pointing out of the page, thereby defining the direction of the positive Z axis. By necessity, the −Z axis is oriented perpendicularly into the page. Using the right-hand rule means that only two axes ever need to be defined and shown; use of the right-hand rule allows the third axis to be completely described.

SPECIAL FOCUS 4-5 Another Use of the “Right-Hand Rule”: a Guide for Describing the Direction of Angular Motion and Torque

Another use of the right-hand rule is to define the rotation direction of angular motion and torque. Consider once again, the coordinate reference frame depicted in Figure 4-7. This reference frame indicates that the path of humeral motion (abduction) is in the X-Y (frontal) plane, around a perpendicular anterior-posterior axis (or, as described in Special Focus 4-4, the Z axis). The right-hand rule is again applied to Figure 4-7 as follows. Begin by aligning the ulnar side of your right hand parallel with the arm segment of the model, so that flexing your fingers curls them in the rotation path of shoulder abduction. The direction of your extended thumb points in the +Z direction, indicating abduction is a positive Z rotation. Shoulder adduction is in a negative Z direction.

This right-hand rule is also used to describe the rotary direction of torque. Again the right hand is used, curling your fingers in the path of motion produced by the torque. Returning to Figure 4-7, force M, produced by the shoulder abductor muscles, generates a +Z torque, whereas the shoulder adductor muscles (not shown) generate a −Z torque. With the coordinate reference frame oriented as shown, the shoulder abductors will always generate a +Z torque, regardless of concentric action (associated with a +Z motion), or eccentric action (associated with a −Z motion).

Rotary or angular movements or torques are often described as occurring in a plane, around a perpendicular axis of rotation. In most kinesiologic literature, a segment’s rotation direction is typically described by terms such as flexion and extension and, to a lesser extent, clockwise or counterclockwise rotation. Such a system is adequate for most clinical analysis and is used throughout this textbook. More formal, quantitative analysis, however, may be necessary to designate the direction of angular motion and torques. Such a system is based on the 3D Cartesian coordinate reference frame and uses another form of the right-hand rule,⁴ as described in Special Focus 4-5.

In closing, analyzing movement within three dimensions is more complicated than in two dimensions, but it does provide a more comprehensive profile of human movement. There are excellent resources available that describe techniques for conducting 3D analysis, and some of these references are provided at the end of the chapter.2,22,23 The quantitative analysis described in this chapter focuses on movements that are restricted to two dimensions.

Forces and Torques

As vector quantities, forces can be analyzed in different manners depending on the context of the analysis. Several forces can be combined into a single resultant force, represented by a single vector. Adding forces together uses processes called vector composition. Alternatively, a single force may be resolved or “decomposed” into two or more forces, the combination of which has the exact effect of the original force. This process of decomposing a single force into its components is termed vector resolution. The analysis of vectors using the processes of composition and resolution provides the means of understanding how forces rotate or translate body segments and subsequently cause rotation, compression, shear, or distraction at the joint surfaces.

GRAPHIC AND MATHEMATIC METHODS OF FORCE ANALYSIS

Composition and resolution of forces can be accomplished using graphic methods of analysis, or mathematic methods including the simple addition and subtraction of vectors or, in some cases, right-angle trigonometry. The graphic method of force analysis represents force or force component vectors as arrows and is performed by aligning them in a tip-to-tail fashion. A drawback to this method is that it requires a high degree of precision drawing. The length of the arrows must be precisely scaled to the magnitudes of the forces, and the orientations and directions of the arrows must match the forces exactly.

The trigonometric method does not require the same precision of drawing, and often provides a more accurate method of force analysis. This method uses rectangular components, and “right-angle trigonometry” to determine magnitudes and angles of forces. The trigonometric functions are based on the relationship that exists between the angles and sides of a right triangle. Refer to Appendix I, Part A, for a brief review of this material.

Proficiency in these techniques is needed to represent and subsequently calculate muscle and joint forces. Both graphic and trigonometric methods are illustrated next, but the remainder of the chapter will use only the trigonometric method.

Composition of Forces: Two or more forces are collinear if they share a common line of force. Vector composition allows several collinear forces to be simply combined graphically as a single resultant force (Figure 4-8). In Figure 4-8, A, the weight of the shank-and-foot segments (S) and the exercise weight (W) are added graphically by means of a ruler and a scale factor determined for the vectors. In this example, S and W act downward, so the resultant force (R) also acts downward and has the tendency to distract (pull apart) the knee joint. R is found graphically by aligning the tail of W to the tip of S. The resultant force R is depicted by the blue arrow that starts at the tail of S and ends at the tip of W. Figure 4-8, B illustrates a cervical traction device that employs a weighted pulley system, acting upward, opposite to the force created by gravity on the center of gravity of the head. Graphically, the tail of H is aligned to the tip of T, and the resultant arrow (R) starts at the tail of T and ends at the tip of H. The upward direction of R (in blue) indicates a net upward distraction force on the head and neck.

FIGURE 4-8. Vector composition of collinear forces. A, Two force vectors are acting on the knee: the weight of the shank-and-foot segment (S) and the exercise weight (W) applied at the ankle. These forces are added to determine the resultant force (R). The X-Y coordinate frame indicates +Y as upward; the negative sign assigned to the forces indicates a downward pull. B, The weight of the head (H) and traction force (T) act along the same line but in opposite directions. R is the algebraic sum of these vectors.

The collinear forces depicted in Figure 4-8 can also be combined by simply adding the force magnitudes of the vectors while paying attention to their directions. In Figure 4-8, A, the coordinate reference frame indicates both S and W are collinear and both acting entirely in a −Y direction. As indicated in the box, the result is found by adding the magnitudes of the collinear forces; in this case the result also acts in a −Y direction. In Figure 4-8, B, the forces are collinear but acting in opposite directions (T acting in a +Y direction, H acting in a −Y direction). Adding the two magnitudes together while paying attention to the direction indicates the result is a 22 N force acting in a +Y direction. In this specific example, a traction force of at least 53 N is needed to offset the weight of the head. Using less force would result in no actual distraction (separation) or the cervical vertebrae. This technique may still, however, provide some therapeutic benefit.

Forces acting on a body may be coplanar (in the same plane), but they may not always be collinear. In this case the individual force vectors may be composed graphically using the polygon method. Figure 4-9 illustrates how the polygon method can be applied to a frontal plane model to estimate the joint reaction force on a prosthetic hip while the subject is standing on one limb. With the arrows drawn in proportion to their magnitude and in the correct orientation, the vectors of body weight (W) and hip abductor muscle force (M) are added in a tip-to-tail fashion (see Figure 4-9, B). The combined effect of the W and M vectors is determined by placing the tail of the M vector to the tip of the W vector. Completing the polygon yields the resultant force (R) starting at the tail of W and traveling to the tip of M. Figure 4-9, B illustrates this process, indicating the magnitude and direction of R. Note that R is equal in magnitude, but opposite in direction, to the prosthetic hip joint reaction force (J) depicted in Figure 4-9, A. An excessively large joint reaction force may, over time, contribute to premature loosening of the hip prosthesis.

FIGURE 4-9. A, Three forces are shown acting on a pelvis that is involved in single-limb standing over a right prosthetic hip joint. The forces are hip abductor muscle force (M), body weight (W), and prosthetic hip joint reaction force (J). B, The polygon (or tip-to-tail) method is used to determine the magnitude and direction of the resultant force (R), based on the magnitude and direction of M and W. J in A is equal in magnitude and opposite in direction to R in B. (From Neumann DA: Hip abductor muscle activity in persons who walk with a hip prosthesis while using a cane and carrying a load, Phys Ther 79:1163, 1999.)

A parallelogram can also be constructed to determine the resultant of two coplanar but noncollinear forces. Instead of placing the force vectors tip to tail, as discussed in the previous example, the resultant vector can be found by drawing a parallelogram based on the magnitude and direction of the two component force vectors. As with all graphic techniques of vector analysis, practice is required to be able to relatively accurately draw the size and orientation of the associated force vectors. Figure 4-10 provides an illustration of the parallelogram method used to combine several component vectors into one resultant vector. The component force vectors, F₁ and F₂ (black solid arrows), are generated by the pull of the flexor digitorum superficialis and profundus as they pass palmar (anterior) to the metacarpophalangeal joint. The diagonal, originating at the intersection of F₁ and F₂, represents the resultant force (R) (see Figure 4-10, thick red arrow). Because of the angle between F₁ and F₂, the resultant force tends to raise the tendons palmarly away from the joint. Clinically, this phenomenon is described as a bowstringing force because of the tendons’ resemblance to a pulled cord connected to the two ends of a bow. Normally the bowstringing force is resisted by forces developed in the flexor pulley and collateral ligaments (see force P in blue in Figure 4-10). In severe cases of rheumatoid arthritis, for example, the bowstringing force may eventually rupture the ligaments and dislocate the metacarpophalangeal joints.

FIGURE 4-10. Parallelogram method is used to illustrate the effect of two force vectors (F₁ and F₂) produced by contraction of the flexor digitorum superficialis and profundus muscles across the metacarpophalangeal (MCP) joint. The resultant force (R) vector creates a bowstringing force resisted by the flexor pulley and collateral ligaments (force P in blue) at the MCP joint.

In summary, when two or more forces applied to a segment are combined into a single resultant force, the magnitude of the resultant force is considered equal to the sum of the component vectors. The resultant force can be determined graphically as summarized in the box.

Summary of How to Graphically Compose Force Vectors

• Collinear force vectors can be combined by simple addition or subtraction (see Figure 4-8).

• Nonparallel, coplanar force vectors can be composed by using the polygon (tip-to-tail) method (see Figure 4-9) or the parallelogram method (see Figure 4-10).

Resolution of Forces: The previous section illustrates the composition method of representing forces, whereby multiple coplanar forces acting on a body are replaced by a single resultant force. In many clinical situations, however knowledge of the effect of the individual components that produce the resultant force may be more relevant to understanding the impact of these forces on motion and joint loading, as well as developing specific treatment strategies. Vector resolution is the process of replacing a single force with two or more forces that when combined are equivalent to the original force.

One of the most useful applications of the resolution of forces involves the description and calculation of the rectangular components of a muscle force. As depicted in Figure 4-11, the rectangular components of the muscle force are shown at right angles to each other and are referred to as the X and Y components (M_X and M_Y). (The X axis is set to be parallel to the long axis of the segment, with positive directed distally.) In the elbow model depicted in Figure 4-11, the X component represents the component of the muscle force that is directed parallel to the forearm. The effect of this force component is to compress and stabilize the joint or, in some cases, distract or separate the segments forming the joint. The X component of a muscle force does not produce a torque when it passes through the axis of rotation because it has no moment arm (see Figure 4-11, M_X). In the model depicted in Figure 4-11, the Y component represents the component of the muscle force that acts perpendicularly to the long axis of the segment. Because of the internal moment arm (see Chapter 1) associated with this force component, one effect of M_Y is to cause a rotation (i.e., produce a torque). In this example, the M_Y component may also create a shear force at the humeroradial joint that tends to cause a translation of the bony segment in the +Y direction.

FIGURE 4-11. The muscle force (M) produced by the brachioradialis is represented as the hypotenuse (diagonal) of the rectangle. The X component (M_X) and the Y component (M_Y) are also indicated. The internal moment arm (IMA) is the perpendicular distance between the axis of rotation (red circle) and M_Y. The X-Y coordinate reference frame is placed so the X axis is parallel with the body segment of interest; the thin black arrowheads point toward positive directions.

For the purposes of this chapter, anatomic joints will be considered as frictionless hinge or pin joints with a stationary axis of rotation, allowing rotation in only one plane. Although it is fully recognized that even the simplest joint in the body is far more complex than this, consideration as pin joints allows a much easier understanding of the concepts of this chapter. For example, if the X component of the muscle force (M_X) is directed toward the elbow joint as in Figure 4-11, it may be assumed that the muscle force causes compression of the radial head against the capitulum of the humerus. The Y component of the muscle force (M_Y in Figure 4-11) causes a shear, tending to move the forearm in the +Y direction (in this case upward and slightly posteriorly). As described later, these forces are opposed by the oppositely directed joint reaction forces. Table 4-4 summarizes the characteristics of the X and Y force components of a muscle, as illustrated in Figure 4-11.

TABLE 4-4.

Typical Characteristics of X and Y Components of a Muscle Force (as Illustrated in Figure 4-11)

Y Muscle Force Component	X Muscle Force Component
Acts perpendicular to a bony segment.	Acts parallel to a bony segment.
Often indicated as M_Y, depending on the choice of the reference system.	Often indicated as M_X, depending on the choice of the reference system.
Can cause translation of the bone and/or torque if moment arm >0.	Can cause translation of the bone. Often does not cause a torque because the chosen reference system reduces the moment arm to zero.
In a simple hinge joint model, M_Y creates a shear force between the articulating surfaces. (In reality, M_Y can create shear, compressive, and distractive forces depending on the anatomic complexity of the joint surfaces.)	In a simple hinge joint model, M_X creates a compression or distraction force between the articulating surfaces. (In reality, M_X can create shear, compressive, and distractive forces depending on the anatomic complexity of the joint surfaces.)

CONTRASTING INTERNAL VERSUS EXTERNAL FORCES AND TORQUES

The previously described examples of resolving forces into X and Y components focused on the forces and torques produced by muscle. As described in Chapter 1, muscles, by definition, produce internal forces and torques. The resolution of forces into X and Y components can also be applied to external forces acting on the human body, such as those from gravity, physical contact, external loads and weights, and manual resistance as applied by a clinician. In the presence of an external moment arm, external forces produce an external torque. Generally, in a condition of equilibrium the external torque acts relative to the joint’s axis of rotation in an opposite rotary direction as the net internal torque.

Figure 4-12 illustrates the resolution of both internal and external forces of an individual who is performing an isometric knee extension exercise. Three forces are depicted in Figure 4-12, A: the internal knee extensor muscle force (M), the external shank-and-foot segment weight (S), and the external exercise weight (W) applied at the ankle. Forces S and W act at the center of their respective masses.

FIGURE 4-12. Resolution of internal forces (red) and external forces (black and green) for an individual performing an isometric knee extension exercise. A, The following resultant force vectors are depicted: muscle force (M) of the knee extensors; shank-and-foot segment weight (S); and exercise weight (W) applied at the ankle. B, The free body diagram shows the forces resolved into their X and Y components. The joint reaction force (J) is also shown (blue). In both A and B, the open circles mark the medial-lateral axis of rotation at the knee. (Vectors are not drawn to scale.) Observe that the X-Y coordinate reference frame is set so the X direction is parallel to the shank segment; thin black arrowheads point toward the positive direction.

Figure 4-12, B shows the free body diagram of the exercise performed in A, with M, S, and W resolved into their X and Y components. Assuming static rotary and linear equilibrium, the governing torque (T) and force (F) equations listed to the left of the figure may be used to solve unknown variables. This topic will be addressed in the final section of the chapter.

INFLUENCE OF CHANGING THE ANGLE OF THE JOINT

The relative magnitude of the X and Y components of internal and external forces applied to a bone depends on the position of the limb segment. Consider first how the change in angular position of a joint alters the angle-of-insertion of the muscle (see glossary, Chapter 1). Figure 4-13 shows the constant magnitude biceps muscle force (M) at four different elbow joint positions, each with a different angle-of-insertion to the forearm (designated as α in each of the four parts of the figure). Note that each angle-of-insertion results in a different combination of M_X and M_Y force components. The M_X component creates compression force if it is directed toward the elbow, as in Figure 4-13, A, or distraction force if it is directed away from the elbow as in Figure 4-13, C and D. By acting with an internal moment arm (brown line labeled IMA), the M_Y components in Figure 4-13, A through D generate a +Z torque (flexion torque) at the elbow.

FIGURE 4-13. Changing the angle of the elbow joint alters the angle-of-insertion of the muscle to the forearm. These changes, in turn, alter the magnitude of the X (M_X) and Y (M_Y) components of the biceps muscle force (M). Using trigonometric functions, the magnitudes of M_X and M_Y can be found for each position: A, angle-of-insertion of 20 degrees; B, angle-of-insertion of 90 degrees; C, angle-of-insertion of 45 degrees; and D, angle-of-insertion of 15 degrees. Although the magnitude of M is assumed to be constant (120 N), the changing magnitude of M_Y alters the internal torque significantly throughout the range of motion. The internal moment arm (IMA) is drawn as a brown line extending from the axis of rotation (black dot) to the point of application of M and remains constant throughout A to D. Note that the X-Y coordinate reference frame is set so the X direction is always parallel to the forearm segment; thin black arrowheads point toward the positive direction. (Modified from LeVeau BF: Williams & Lissner’s biomechanics of human motion, ed 3, Philadelphia, 1992, Saunders.)

As shown in Figure 4-13, A, a relatively small angle-of-insertion of 20 degrees favors a relatively large X component force, which directs a larger percentage of the total muscle force to compress the joint surfaces of the elbow. Because the angle-of-insertion is less than 45 degrees in Figure 4-13, A, the magnitude of the M_X component exceeds the magnitude of the M_Y component. When the angle-of-insertion of the muscle is 90 degrees (as in Figure 4-13, B), 100% of M is in the Y direction and is available to produce an elbow flexion torque. At an angle-of-insertion of 45 degrees (Figure 4-13, C), the M_X and M_Y components have equal magnitude, with each about 71% of M. In Figure 4-13, C and D, the angle-of-insertion (shown to the right of M as α) produces a M_X component that is directed away from the joint, thereby producing a distracting or separating force on the joint.

In Figure 4-13, A through D, the internal torque is always in a +Z direction and is the product of M_Y and the internal moment arm (IMA). Even though the magnitude of M is assumed to remain constant throughout the range of motion, the change in M_Y (resulting from changes in angle-of-insertion) produces differing magnitudes of internal torque. Note that the +Z (flexion) torque ranges from 0.93 Nm at near full elbow flexion to 3.60 Nm at 90 degrees of elbow flexion—a near fourfold difference. This concept helps explain why people have greater strength (torque) in certain parts of the joint’s range of motion. The torque-generating capabilities of the muscle depend not only on the angle-of-insertion, and subsequent magnitude of M_Y, but also on other physiologic factors, discussed in Chapter 3. These include muscle length, type of activation (i.e., isometric, concentric, or eccentric), and velocity of shortening or elongation of the activated muscle.

Changes in joint angle also can affect the amount of external or “resistance” torque encountered during an exercise. Returning to the example of the isometric knee extension exercise, Figure 4-14 shows how a change in knee joint angle affects the Y component of the external forces S and W. The external torque generated by gravity on the segment (S) and the exercise weight (W) is equal to the product of the external moment arm (brown line labeled EMA in B and C) and the Y component of the external forces (S_Y and W_Y). In Figure 4-14, A, no external torque exists in the sagittal plane because S and W force vectors are entirely in the +X direction (S_Y and W_Y = 0). The S and W vectors are directed through the knee’s axis of rotation and therefore have no external moment arm. Because these external forces are pointed in the +X direction, they tend to distract the joint. Figure 4-14, B and C show how a greater external torque is generated with the knee fully extended (in C) compared with the knee flexed 45 degrees (in B). Although the magnitude of the external forces, S and W, are the same in all three cases, the −Z directed (flexion) external torque is greatest when the knee is in full extension. As a general principle, the external torque around a joint is greatest when the resultant external force vector intersects the bone or body segment at a right angle (as in Figure 4-14, C). When free weights are used, for example, external torque is generated by gravity acting vertically. Resistance torque from the weight is therefore greatest when the body segment is positioned horizontally. Alternatively, with use of a cable attached to a column of stacked weights, resistance torque from the cable is greatest in the position where the cable acts at a right angle to the segment. Note that this is often in a different position than where the torque caused by gravity acting on the segment is greatest. Resistive elastic bands and tubes present further complications, as resistance torque from these devices varies with the angle of the resistance force vector and the amount of stretch in the device; both factors vary through a range of motion.^19,21

FIGURE 4-14. A change in knee joint angle affects the magnitude of the components of the external forces generated by the shank-and-foot segment weight (S) and exercise weight (W). In A, all of W and S act in the +X direction and have no external moment arms to produce a sagittal plane external torque at the knee. In B and C, S_Y and W_Y act in a −Y direction, and each possesses an external moment arm ( is equal to the external moment arm for S_Y; is equal to the external moment arm for W_Y). Different external torques are generated at each of the three knee angles. The X-Y coordinate reference frame is set so the X direction is parallel to the shank segment; thin black arrowheads point toward the positive direction.

SPECIAL FOCUS 4-6 Designing Resistive Exercises So That the External and Internal Torque Potentials Are Optimally Matched

The concept of altering the angle of a joint is frequently used in exercise programs to adjust the magnitude of resistance experienced by the patient or client. It is often desirable to design an exercise program so that the external torque matches the internal torque potential of the muscle or muscle group. Consider a person performing a “biceps curl” exercise, shown in Figure 4-17, A. With the elbow flexed to 90 degrees, both the internal and external torque potentials are greatest, because the product of each resultant force (M and W) and their moment arms (IMA and EMA) are maximal. At this elbow position the internal and external torque potentials are maximal as well as optimally matched. As the elbow position is altered in Figure 4-17, B, the external torque remains the same; however, the angle-of-insertion of the muscle is different, requiring a much larger muscle force, M, to produce the same internal +Z directed torque. Note the Y component of the muscle force (M_Y) in Figure 4-17, B has the same magnitude as the muscle force M in Figure 4-17, A. A person with significant weakness of the elbow flexor muscle may have difficulty holding an object in position B but may have no difficulty holding the same object in position A.

FIGURE 4-17. Changing the angle of elbow flexion can alter both internal and external torque potential. A, The 90-degree position of the elbow maximizes the potential for both the internal and the external torque. B, With the forearm horizontal and the elbow closer to extension, the external torque remains maximal but the overall biceps force (M) must increase substantially to yield sufficient M_Y force to support the weight. EMA, external moment arm; IMA, internal moment arm; M, muscle force; M_Y, Y component of the muscle force; W, exercise weight.

COMPARING TWO METHODS FOR DETERMINING TORQUE AROUND A JOINT

In the context of kinesiology, a torque is the effect of a force tending to rotate a body segment around a joint’s axis of rotation. Torque is the rotary equivalent of a force. Mathematically, torque is the product of a force and its moment arm and usually is expressed in units of newton-meters. Torque is a vector quantity, having both magnitude and direction.

Two methods for determining torque yield identical mathematic solutions. Understanding both methods provides valuable insight into the concept of torque, especially how it relates to clinical kinesiology. The methods apply to both internal and external torque, assuming that the system in question is in rotational equilibrium (i.e., the angular acceleration around the joint is zero).

Internal Torque: The two methods for determining internal torque are illustrated in Figure 4-15. Method 1 calculates the internal torque as the product of M_Y and its internal moment arm (). Method 2 uses the entire muscle force (M) and therefore does not require this variable to be resolved into its rectangular components. In this method, internal torque is calculated as the product of the muscle force (the whole force, not a component) and IMA_M (i.e., the internal moment arm that extends perpendicularly between the axis of rotation and the line of action of M). Methods 1 and 2 yield the same internal torque because they both satisfy the definition of a torque (i.e., the product of a force and its associated moment arm). The associated force and moment arm for any given torque must intersect each other at a 90-degree angle.

FIGURE 4-15. The internal (muscle-produced) flexion torque at the elbow can be determined using two different methods. Method 1 calculates torque as the product of the Y component of the muscle force (M_Y) times its internal moment arm (). Method 2 calculates torque as the product of the entire force of the muscle (M) times its internal moment arm (IMA_M). Both expressions yield equivalent internal torques. The axis of rotation is depicted as the open black circle at the elbow. The X-Y coordinate reference frame is set so the positive X direction is parallel to the forearm segment.

External Torque: Figure 4-16 shows an external torque applied to the elbow through a resistance produced by an elastic band (depicted in green as R). The weight of the body segment is ignored in this example. Method 1 determines external torque as the product of R_Y times its external moment arm (). Method 2 uses the product of the band’s entire resistive force (R) and its external moment arm (EMA_R). As with internal torque, both methods yield the same external torque because both satisfy the definition of a torque (i.e., the product of a resistance [external] force and its associated external moment arm). The associated force and moment arm for any given torque must intersect each other at a 90-degree angle.

FIGURE 4-16. An external torque is applied to the elbow through a resistance generated by tension in a cable (R). The weight of the body segment is ignored. The external torque can be determined using two different methods. Method 1 uses the product of the Y component of the resistance (R_Y) times its external moment arm (). Method 2 calculates torque as product of the entire force of the resistance (R) times its external moment arm (EMA_R). Both expressions yield equivalent external torques. The axis of rotation is depicted as the open black circle at the elbow. The X-Y coordinate reference frame is set so the positive X direction is parallel to the forearm segment.

MANUALLY APPLYING EXTERNAL TORQUES DURING EXERCISE AND STRENGTH TESTING

External or resistance torques are often applied manually during an exercise program. For example, if a patient is beginning a knee rehabilitation program to strengthen the quadriceps muscle, the clinician may initially apply manual resistance to the knee extensors at the midtibial region. As the patient’s knee strength increases, the clinician can exert a greater force at the midtibial region, or the same force near the ankle.

Because external torque is the product of a force (resistance) and an associated external moment arm, an equivalent external torque can be applied by a relatively short external moment arm and a large external force, or a long external moment arm and a smaller external force. The knee extension resistance exercise depicted in Figure 4-18 shows that the same external torque (15 Nm) can be generated by two combinations of external forces and moment arms. Note that the resistance force applied to the leg is greater in Figure 4-18, A than in Figure 4-18, B. The higher contact force may be uncomfortable for the patient, and this factor needs to be considered during the application of resistance. A larger external moment arm, as shown in Figure 4-18, B, may be necessary if the clinician chooses to manually challenge a muscle group as potentially forceful as the quadriceps. Even using a long external moment arm, clinicians may be unable to provide enough torque to maximally resist large and strong muscle groups.¹¹

FIGURE 4-18. The same external torque (15 Nm) is applied against the quadriceps muscle by using a relatively large resistance (100 N) and small external moment arm (A), or a relatively small resistance (50 N) and large external moment arm (B). The external moment arms are indicated by the brown lines that extend from the medial-lateral axis of rotation at the knee.

A hand-held dynamometer is a device used to manually measure the maximal isometric strength of certain muscle groups. This device directly measures the force generated between the device and the limb during a maximal-effort muscle contraction. Figure 4-19 shows this device used to measure the maximal-effort, isometric elbow extension torque in an adult woman. The external force (R) measured by the dynamometer is in response to the internal force generated by the elbow extensor muscles (E). Because the test is performed isometrically, the measured external torque (R × EMA) will be equal in magnitude but opposite in direction to the actively generated internal torque (E × IMA). If the clinician is documenting external force (as indicated by the dial on the dynamometer), he or she needs to pay close attention to the position of the dynamometer relative to the person’s limb. Changing the external moment arm of the device will alter the external force reading. This is shown by comparing the two placements of the dynamometer in Figure 4-19, A and B. The same elbow extension internal force (E) will result in two different external force readings (R). The longer external moment arm used in Figure 4-19, A, results in a lower external force than the shorter external moment arm used in Figure 4-19, B. On repeated testing, for example, before and after a strengthening program, the force dynamometer must be positioned with exactly the same external moment arm to allow a valid strength comparison to the prestrengthening values. Documenting external torques rather than forces does not require the external moment arm to be exactly the same for every testing session. The external moment arm does need to be measured each time, however, to allow conversion of the external force (as measured by the force dynamometer) to external torque (the product of the external force and the external moment arm).

FIGURE 4-19. A dynamometer is used to measure the maximal, isometric strength of the elbow extensor muscles. The external moment arm (EMA) is the distance between the axis of rotation (open circle) and the point of external force (R) measured by the dynamometer. The different placement of the device on the limb creates different EMAs in A and B. Elbow extension force (E), which is the same in A and B, generates equivalent internal torques through its internal moment arm (IMA), which is equal in magnitude but opposite in direction to the external torques generated by the product of R and EMA. The joint reaction force (J), shown in blue, is equal but opposite in direction to the sum of R + E. The distal application of the measuring device shown in A results in a longer EMA and a lower external force reading. Because R is less, J is also less. The more proximal application of the device in B results in a shorter EMA, a higher external force reading, and a greater J. Vectors are drawn to approximate scale. (The X-Y coordinate reference frame is set so the X direction is parallel to the forearm; thin black arrowheads point in positive directions. Based on conventions described in the next section [summarized in Box 4-1], the internal moment arm is assigned a negative number. This, in turn, appropriately assigns opposite rotational directions to the opposing torques.)

Note also that although the elbow extension internal force and torque are the same in Figure 4-19, A and B, the joint reaction force (J) and external force (R) are higher in Figure 4-19, B. This means that the pressure between the force dynamometer pad and the patient’s skin is higher and could potentially cause discomfort. In some cases the discomfort could be great enough to reduce the amount of internal torque the patient is willing to develop, thereby influencing a maximal strength assessment. In addition, a higher magnitude of joint reaction force could have implications in conditions of compromised articular cartilage.

INTRODUCTION TO BIOMECHANICS: FINDING THE SOLUTIONS

In the previous sections, concepts were introduced that provide the framework for quantitative methods of biomechanical analysis. Many approaches are applied when solving problems in biomechanics. These approaches can be employed to assess (1) the effect of a force at an instant in time (force-acceleration relationship); (2) the effect of a force applied over an interval of time (impulse-momentum relationship); and (3) the application of a force that causes an object to move through some distance (work-energy relationship). The particular approach selected depends on the objective of the analysis. The subsequent sections in this chapter are directed toward the analysis of forces or torques at one instant in time, or the force (torque)-acceleration approach.

When one considers the effects of a force, or forces, and the resultant acceleration at an instant in time, two situations can be defined. In the first case the effects of the forces cancel and there is no acceleration because the object is either stationary or moving at a constant velocity. This is the situation described previously as equilibrium and is analyzed using a branch of mechanics known as statics. In the second situation, linear and/or angular acceleration is occurring because the system is subjected to unbalanced forces or torques. In this situation the system is not in equilibrium, and analysis requires using a branch of mechanics known as dynamics. Static analysis is the simpler approach to problem solving in biomechanics and is the focus of this chapter. Although clinicians do not often mathematically complete the types of analyses contained in this chapter, a full appreciation of the biomechanics of normal and abnormal motion, including most treatment techniques, is facilitated through learning the components of the mathematic analysis. For example, recommendations for treatment of articular cartilage disorders are better made with consideration of the variables that influence compressive joint reaction force. Ligament reconstruction grafts often require a period of protective loading; this can be safely accomplished while strengthening muscles only if the magnitude and direction of muscle and joint forces are considered. The reader is encouraged to consider these types of clinical issues by answering the questions posed at the completion of each of the upcoming three sample problems.

Static Analysis

Biomechanical studies often induce conditions of static equilibrium in order to simplify the approach to the analysis of human movement. In static analysis the system is in equilibrium because it is not experiencing acceleration. As a consequence the sum of the forces and the sum of the torques acting on the system are zero. The forces and torques in any direction are completely balanced by the forces and torques acting in the opposite direction. Because in static equilibrium there is no linear or angular acceleration, the inertial effect of the mass and moment of inertia of the bodies can be ignored.

The force equilibrium equations, Equations 4.13 A and B, are used for static (uniplanar) translational equilibrium. In the case of static rotational equilibrium, the sum of the torques around any axis of rotation is zero. The torque equilibrium equation, Equation 4.14, is also included. The equations previously depicted in Figure 4-19 provide a simplified example of static rotational equilibrium about the elbow. The muscle force of the elbow extensors (E) times the internal moment arm (IMA) creates a potential extension (clockwise, −Z) torque. This torque (product of E and IMA) is balanced by a flexion (counterclockwise, +Z) torque provided by the product of the transducer’s force (R) and its external moment arm (EMA). Assuming no movement of the elbow, ΣT_Z = 0; in other words, the opposing torques at the elbow are assumed to be equal in magnitude and opposite in direction.

Governing Equations for Static Uniplanar Analysis

Forces and Torques Are Balanced

Force (F) Equilibrium Equations

(Equation 4.13 A)

(Equation 4.13 B)

Torque (T) Equilibrium Equation

(Equation 4.14)

GUIDELINES FOR PROBLEM SOLVING

The guidelines listed in Box 4-1 are necessary to follow the logic for solving the upcoming three sample problems. (Although most concepts listed in Box 4-1 have been described previously in this chapter, guideline 5 is new. This particular guideline describes the convention used to assign direction to moment arms.) In each of the three upcoming problems, an assumption of static equilibrium is required to solve the magnitude and direction of torque, muscle force, and joint reaction force.

BOX 4-1. Guidelines for Solving for Muscle Force, Torque, and Joint Reaction Force

1. Draw the free body diagram, isolating the body segment(s) under consideration. Draw in all forces acting on the free body, including, if appropriate, gravity, resistance, muscular, and joint reaction forces. Identify the axis of rotation at the center of the joint.

2. Establish an X-Y reference frame that will specify the desired orientation of the X and Y components of forces. Designate the X axis parallel with the isolated body segment (typically a long bone), positive pointing distally. The Y axis is oriented perpendicularly to the same body segment. (Use arrowheads on the X and Y axes to designate positive directions.)

3. Resolve all known forces into their X and Y components.

4. Identify the moment arms associated with each Y force component. The moment arm associated with a given Y force component is the perpendicular distance between the axis of rotation and the line of force. Note that joint reaction force and the X components of all forces will not have a moment arm, because the line of force of these forces typically passes through the axis of rotation (center of the joint).

5. Assign a direction to the moment arms. By convention, moment arms are measured from the axis of rotation to the Y component of the force. If this measurement travels in a positive X direction, it is assigned a positive value. If the measurement travels in a negative X direction, it is assigned a negative value.

6. Use ΣT_Z = 0 (Equation 4.14) to find the unknown muscle torque and force.

7. Use ΣF_X = 0 and ΣF_Y = 0 (Equations 4.13 A and B) to find the X and Y components of the unknown joint reaction force.

8. Compose X and Y components of the joint reaction force to find the magnitude of the total joint reaction force.

Note: There are other, more elegant methods to determine torques and component forces in systems similar to those illustrated in this chapter. However, these methods require a working knowledge of cross products, dot products, and unit vectors, topics that are beyond the scope of this chapter.

Additional problem-solving examples and related clinical questions are available in Appendix I, Part B.

Problem 1: Consider the situation posed in Figure 4-20, A, in which a person generates an isometric elbow flexor muscle force at the elbow while holding a weight in the hand. Assuming equilibrium, the three unknown variables are the (1) internal (muscle-produced) elbow flexion torque, (2) elbow flexor muscle force, and (3) joint reaction force at the elbow. All abbreviations and pertinent data are included in the box associated with Figure 4-20.

FIGURE 4-20. Problem 1. A, An isometric elbow flexion exercise is performed against an exercise weight held in the hand. The black dot marks the segment’s center of gravity; the exercise weight’s center of gravity is marked by the green dot. The forearm is held in the horizontal position. B, A free body diagram is shown of the exercise, including a box with the abbreviations and data required to solve the problem. The medial-lateral axis of rotation at the elbow is shown as an open red circle; the vectors are not drawn to scale. (The X-Y coordinate reference frame is set so the X direction is parallel to the forearm; black arrowheads point in positive directions.)

To begin, a free body diagram and X-Y reference frame is constructed (see Figure 4-20, B). The axis of rotation and all moment arm distances are indicated. Although at this point the direction of the joint (reaction) force (J) is unknown, it is assumed to act in a direction opposite to the pull of muscle. This assumption generally holds true in an analysis in which the mechanical advantage of the system is less than one (i.e., when the muscle forces are greater than the external resistance forces) (see Chapter 1).

Resolving Known Forces into X and Y Components: In the elbow position depicted in Figure 4-20, all forces act parallel to the Y axis; there is no force acting in the X direction. This means the magnitude of the Y components of the forces is equal to the magnitude of the entire force, and the X components are all zero. This situation is unique to this position, in which muscle force and gravity are vertical and the segment is positioned horizontally.

The magnitude of the forces are determined through trigonometric functions, then the direction (+ or −) is applied.

Solving for Internal Torque and Muscle Force: The external torques originating from the weight of the forearm-hand segment (S_Y) and the exercise weight (W_Y) generate a −Z (clockwise, extension) torque about the elbow. In order for the system to remain in equilibrium, the elbow flexor muscles have to generate an opposing internal +Z (counterclockwise, flexion) torque. Summing the torques around the elbow axis allows the line-of-action of J to pass through the axis, thus making the moment arm of J equal to zero. This results in only one unknown in Equation 4.14: the magnitude of the muscle force:

The resultant muscle (internal) force is the result of all the active muscles that flex the elbow. This type of analysis does not, however, provide information about how the force is distributed among the various elbow flexor muscles. This requires more sophisticated procedures, such as muscle modeling and optimization techniques, which are beyond the scope of this text.

The magnitude of the muscle force is over six times greater than the magnitude of the external forces (i.e., forearm-hand weight and load weight). The larger force requirement can be explained by the disparity in moment arm length of the elbow flexors when compared with the moment arm lengths of the two external forces. The disparity in moment arm lengths is not unique to the elbow flexion model, but it is ubiquitous throughout the muscular-joint systems in the body. For this reason, most muscles of the body routinely generate force many times greater than the externally applied force. The combinations of external and muscular forces often require bone and articular cartilage to absorb and transmit very large joint forces, sometimes resulting from seemingly nonstressful activities. The next set of calculations determines the magnitude and direction of the joint reaction force.

Solving for Joint Reaction Force: Because the joint reaction force (J) is the only remaining unknown variable depicted in Figure 4-20, B, this variable is determined by Equations 4.13 A and B.

Because there are no X components of M or either of the two external forces, the joint reaction force does not have an X component either.

The negative Y component of the joint reaction force indicates that the joint force acts in a −Y direction (downward).

Total joint reaction force can be found by using the Pythagorean theorem with the X and Y components. (This step may not be necessary for problems such as this, where one of the component forces is zero, but it is included here for consistency of method.)

Because muscle force is usually the largest force acting about a joint, the direction of the net joint reaction force often opposes the pull of the muscle. Without such a force, for example, the muscle indicated in Figure 4-20 would accelerate the forearm upward, resulting in an unstable joint. In short, the joint reaction force in this case (largely supplied by the humerus pushing against the trochlear notch of the ulna) provides the required force to maintain linear static equilibrium at the elbow. As stated earlier, the joint reaction force does not produce a torque because it is assumed to act through the axis of rotation and therefore has a zero moment arm.

Clinical Questions Related to Problem 1:

1. Assume a patient with osteoarthritis of the elbow is holding a load similar to that depicted in Figure 4-20. How would you respond to the question posed by a patient, “Why would my elbow be so painful from holding such a light weight?”

2. Describe a few clinical conditions in which the magnitude and direction of the joint reaction force could be biomechanically (physiologically) unhealthy for a patient.

3. Which variable is most responsible for the magnitude and direction of the joint reaction force at the elbow?

4. Assume a person with a recent elbow joint replacement needs to strengthen the elbow flexor muscles. Given the isometric situation depicted in Figure 4-20:

a. How could the joint reaction force on the elbow be reduced while the same size exercise weight is used?

b. How could the joint reaction force on the elbow be reduced while the same magnitude of external torque is created?

Answers to the clinical questions can be found on the Evolve website.

Problem 2: In Problem 1 the forearm is held horizontally, thereby orienting the internal and external forces perpendicular to the forearm. Although this presentation greatly simplifies the calculations, it does not represent a typical biomechanical situation. Problem 2 shows a more common situation, in which the forearm is held at a position other than the horizontal (Figure 4-21, A). As a result of the change in elbow angle, the angle-of-insertion of the elbow flexor muscles and the angle of application of the external forces are no longer right angles. In principle, all other aspects of this problem are identical to Problem 1. Assuming equilibrium, three unknown variables are once again to be determined: (1) the internal (muscular-produced) torque, (2) the muscle force, and (3) the joint reaction force at the elbow.

FIGURE 4-21. Problem 2. A, An isometric elbow flexion exercise is performed against an identical weight as that depicted in Figure 4-20. The forearm is held 30 degrees below the horizontal position. B, A free body diagram is shown, including a box with the abbreviations and data required to solve the problem. The vectors are not drawn to scale. C, The joint reaction force (J) vectors are shown in response to the biomechanics depicted in B. The X-Y coordinate reference frame is set so the X direction is parallel to the forearm; black arrowheads point in positive directions.

Figure 4-21, B illustrates the free body diagram of the forearm and hand segment held at 30 degrees below the horizontal (θ). The reference frame is established such that the X axis is parallel to the forearm-hand segment, positive pointed distally. All forces acting on the system are indicated, and each is resolved into their respective X and Y components. The angle-of-insertion of the elbow flexors to the forearm (α) is 60 degrees. All numeric data and abbreviations are listed in the box associated with Figure 4-21.

Resolving Known Forces into X and Y Components: Magnitudes of external forces are found through the use of trigonometric functions, then directions (+ or −) are applied based on the established X-Y axis reference frame:

Solving for Internal Torque and Muscle Force:

Because an internal moment arm length of 0.05 m was used, the last calculation yielded the magnitude of its associated perpendicular vector, the Y component of M (M_Y), not the total muscle force M. The total muscle force M is determined as follows:

The X component of the muscle force, M_X, can be solved as follows:

The negative sign was added to indicate M_X is pointed in the −X direction.

Solving for Joint Reaction Force: The joint reaction force (J) and its X and Y components (J_Y and J_X) are shown separately in Figure 4-21, C. (This is done to increase the clarity of the illustration.) The directions of J_Y and J_X are assumed to act generally downward (negative Y) and to the right (positive X), respectively. These are directions that oppose the force of the muscle. The components (J_Y and J_X) of the joint force (J) can be readily determined by using Equations 4.13 A and B, as follows:

As depicted in Figure 4-21, C, J_Y and J_X act in directions that oppose the force of the muscle (M). This reflects the fact that muscle force, by far, is the largest of all the forces acting on the forearm-hand segment. J_X being positive indicates that the joint is under compression, whereas J_Y being negative indicates that the joint is under anterior and superior shear. In other words, if J_Y did not exist, the forearm would accelerate in an anterior and superior (+Y) direction.

The magnitude of the resultant joint force (J) can be determined using the Pythagorean theorem, as follows:

Another characteristic of the joint reaction force that is of interest is the direction of J with respect to the axis of the forearm (X axis). This can be calculated using the inverse cosine function as follows:

The resultant joint reaction force has a magnitude of 394.1 N and is directed toward the joint at an angle of 60 degrees to the forearm segment (i.e., the X axis). It is no coincidence that the angle of approach of J is the same as the angle-of-insertion of the elbow flexor muscles.

Clinical Questions Related to Problem 2:

1. Assume the forearm (depicted in Figure 4-21) is held 30 degrees above rather than below the horizontal plane.

a. Does the change in forearm angle alter magnitude of the external torque?

b. Can you conclude that is it “easier” to hold the forearm 30 degrees above as compared with below the horizontal plane?

2. In what situation would a large force demand on the muscle be a clinical concern?

3. What would happen if, from the position depicted in Figure 4-21, A, the muscle force suddenly decreased or increased slightly?

Answers to the clinical questions can be found on the Evolve website.

Problem 3: Although the forearm was not positioned horizontally in Problem 2, all resultant forces were depicted as parallel. Problem 3 is complicated slightly by the forces not being parallel, and the bony lever system being a first-class (versus a third-class) lever (see Chapter 1). Problem 3 analyzes the isometric phase of a standing triceps-strengthening exercise using resistance applied by a cable (Figure 4-22, A). The patient can extend and hold her elbow partially flexed against the cable transmitting 15 pounds of force from the stack of weights. Assuming equilibrium, three unknown variables are once again to be determined using the same steps as before: (1) the internal (muscular-produced) torque, (2) the muscle force, and (3) the joint reaction force at the elbow.

FIGURE 4-22. Problem 3. A, A standing isometric elbow extension exercise is performed against resistance provided by a cable. The forearm is held 25 degrees from the vertical position. B, A free body diagram is shown, including a box with the abbreviations and data required to solve the problem. The vectors are not drawn to scale. C, The joint reaction force (J) vectors are shown in response to the biomechanics depicted in B. (The X-Y coordinate reference frame is set so the X direction is parallel with the forearm; black arrowheads point in positive directions.)

Figure 4-22, B illustrates the free body diagram of the elbow held partially flexed, with the forearm oriented 25 degrees from the vertical (θ). The coordinate reference frame is again established such that the X axis is parallel to the forearm-hand segment, positive pointed distally. All forces acting on the system are indicated, and each is resolved into their respective X and Y components. The angle-of-insertion of the elbow extensors to the forearm (α) is 20 degrees, and the angle between the cable and the long axis of the forearm (β) is 70 degrees. All numeric data and abbreviations are listed in the box associated with Figure 4-22.

Resolving Known Forces into X and Y Components: Magnitudes of forces are found through the use of trigonometric functions, then directions (+ or −) are applied as in previous problems, as follows:

Solving for Internal Torque and Muscle Force: This system is a first-class lever with the muscle force located on the opposite side of the elbow axis as the external forces. The internal moment arm IMA (as applied to M_Y) is assigned a negative value because the measurement of IMA from the axis of rotation to M_Y travels in a negative X direction (review no. 5 in Box 4-1).

This relatively large Y component of M is necessary because of the small IMA and the large external torque produced by C. The total muscle force, or M, is determined as follows:

The X component of the muscle force, M_X, can be solved as follows:

The negative sign was added to indicate M_X is pointed in the −X direction.

Solving for Joint Reaction Force: The joint reaction force (J) and its X and Y components (J_Y and J_X) are shown separately in Figure 4-22, C. (This is done to increase the clarity of the illustration.) The directions of J_Y and J_X are assumed to act in −Y and +X directions, respectively. These directions oppose the Y and X components of the muscle force. This assumption can be verified by determining the J_Y and J_X components using Equations 4.13 A and B.

As depicted in Figure 4-22, C, J_Y and J_X act in directions that oppose the force of the muscle. J_X being positive indicates that the joint is under compression, whereas J_Y being negative indicates that the joint is experiencing anterior shear. In other words, if J_Y did not exist, the forearm would accelerate in a general anterior (+Y) direction.

The magnitude of the resultant joint force (J) can be determined using the Pythagorean theorem:

Another important characteristic of the joint reaction force is the direction of J with respect to the axis of the forearm (the X axis). This can be calculated using the inverse cosine function:

The resultant joint reaction force has a magnitude of 2879.57 N and is directed toward the elbow at an angle of over 21 degrees to the forearm segment (i.e., the X axis). The angle is almost the same as the angle-of-insertion of the muscle force (α), and the magnitude of J is similar to the magnitude of M. These similarities serve as a reminder of the dominant role of muscle in determining both the magnitude and direction of the joint reaction force. Note that if the M and J vector arrows were drawn to scale with the length of S, they would extend far beyond the limits of the page!

Clinical Questions Related to Problem 3:

1. Figure 4-22 shows the pulley used by the resistance cable located at eye level. Assuming the subject maintains the same position of her upper extremity, what would happen to the required muscle force and components of the joint reaction force if the pulley was relocated at:

a. Chest level?

b. Floor level?

2. How would the exercise change if the pulley was located at floor level with the patient facing away from the pulley?

3. Note in Figure 4-22 that the angle (β) between the force in the cable (C) and the forearm is 70 degrees.

a. At what angle of β would force C produce the greatest external torque?

b. With the pulley at eye level, at what elbow angle would force C produce the greatest external torque?

Answers to the clinical questions can be found on the Evolve website.

Dynamic Analysis

Static analysis is the most basic approach to kinetic analysis of human movement. This form of analysis is used to evaluate forces and torques on a body when there are little or no significant linear or angular accelerations. External forces that act against a body at rest can be measured directly by various instruments, such as force transducers (shown in Figure 4-19), cable tensiometers, and force plates. Forces acting internal to the body are usually measured indirectly by knowledge of external torques and internal moment arms. This approach was highlighted in the previous three sample problems. In contrast, when linear or angular accelerations occur, a dynamic analysis must be undertaken. Walking is an example of a dynamic movement caused by unbalanced forces acting on the body; body segments are constantly speeding up or slowing down, and the body is in a continual state of losing and regaining balance with each step. A dynamic analysis therefore is required to calculate the forces and torques produced by or on the body during walking.

Solving for forces and torques under dynamic conditions requires knowledge of mass, mass moments of inertia, and linear and angular accelerations (for 2D dynamic analysis, see Equations 4.15 and 4.16). Anthropometric data provide the inertial characteristics of body segments (mass, mass moment of inertia), as well as the lengths of body segments and location of axis of rotation at joints. Kinematic data, such as displacement, velocity, and acceleration of segments, are measured through various laboratory techniques, which are described next.^2,18,20,22 This is followed by a description of the techniques commonly used to directly measure external forces, which may be used in static or dynamic analysis.

Two-Dimensional Dynamic Analyses of Force and Torque

KINEMATIC MEASUREMENT SYSTEMS

Detailed analysis of movement requires a careful and objective evaluation of the motion of the joints and body as a whole. This analysis most frequently includes an assessment of position, displacement, velocity, and acceleration. Kinematic analysis may be used to assess the quality and quantity of motion of the body and its segments, the results of which describe the effects of internal and external forces and torques. Kinematic analysis can be performed in a variety of environments, including sport, ergonomics, and rehabilitation. There are several methods to objectively measure human motion, including electrogoniometry, accelerometry, imaging techniques, and electromagnetic tracking devices.

Electrogoniometer: An electrogoniometer measures joint angular rotation during movement. The device typically consists of an electrical potentiometer built into the pivot point (hinge) of two rigid arms. Rotation of a calibrated potentiometer measures the angular position of the joint. The related output voltage is typically measured by a computer data acquisition system. The arms of the electrogoniometer are strapped to the body segments, such that the axis of rotation of the goniometer is approximately aligned with the joint’s axis of rotation (Figure 4-23). The position data obtained from the electrogoniometer combined with the time data can be mathematically converted to angular velocity and acceleration. Although the electrogoniometer provides a fairly inexpensive and direct means of capturing joint angular displacement, it encumbers the subject and is difficult to fit and secure over fatty and muscle tissues. In addition, a uniaxial electrogoniometer is limited to measuring range of motion in one plane. As shown in Figure 4-23, the uniaxial electrogoniometer can measure knee flexion and extension but is unable to detect the subtle but important rotation that also can occur in the horizontal plane. Other types of electrogoniometers exist. Figure 4-24 shows a different style that measures motion in two planes with sensors held onto the subject’s skin by double-sided tape.

FIGURE 4-23. An electrogoniometer is shown strapped to the thigh and leg. The axis of the goniometer contains the potentiometer and is aligned over the medial-lateral axis of rotation at the knee joint. This particular instrument records a single plane of motion of the knee only.

FIGURE 4-24. A biaxial electrogoniometer measuring wrist flexion and extension as well as radial and ulnar deviation. (Courtesy Biometrics, Ltd, Ladysmith, Virginia.)

Accelerometer: An accelerometer is a device that measures acceleration of the object to which it is attached—either an individual segment or the whole body. Linear and angular accelerometers exist but measure accelerations only along a specific line or around a specific axis. Similar to electrogoniometers, multiple accelerometers are required for 3D or multi-segmental analyses. Data from accelerometers are used with body segment inertial information such as mass and mass moment of inertia to estimate net internal forces (F = m × a) and torques (T = I × α). Whole-body accelerometers can be used to estimate an individual’s relative physical activity during daily life.5,6,9

Imaging Techniques: Imaging techniques are the most widely used methods for collecting data on human motion. Many different types of imaging systems are available. This discussion is limited to the systems listed in the box.

Unlike electrogoniometry and accelerometry, which measure movement directly from a body, imaging methods typically require additional signal conditioning, processing, and interpreting before meaningful output is obtained.

Photography is one of the oldest techniques for obtaining kinematic data. With the camera shutter held open, light from a flashing strobe can be used to track the location of reflective markers worn on the skin of a moving subject (see example in Chapter 15 and Figure 15-3). If the frequency of the strobe light is known, displacement data can be converted to velocity and acceleration data. In addition to using a strobe as an interrupted light source, a camera can use a constant light source and take multiple film or digital exposures of a moving event.

Cinematography, the art of movie photography, was once the most popular method of recording motion. High-speed cinematography, using 16-mm film, allowed for the measurement of fast movements. With the shutter speed known, a labor-intensive, frame-by-frame digital analysis on the movement in question was performed. Digital analysis was performed on movement of anatomic landmarks or of markers worn by subjects. Two-dimensional movement analysis was performed with the aid of one camera; 3D analysis, however, required two or more cameras.

For the most part, still photography and cinematography analysis are rarely used today for the study of human motion. The methods are not practical because of the substantial time required for manually analyzing the data. Digital videography has replaced these systems and is one of the most popular methods for collecting kinematic information in both clinical and laboratory settings. The system typically consists of one or more digital video cameras, a signal processing device, a calibration device, and a computer. The procedures involved in video-based systems typically require markers to be attached to a subject at selected anatomic landmarks. Markers are considered passive if they are not connected to another electronic device or power source. Passive markers serve as a light source by reflecting the light back to the camera (Figure 4-25, A). Two-dimensional and 3D coordinates of markers are identified in space by a computer and are then used to reconstruct the image (or stick figure) for subsequent kinematic analysis (see Figure 4-25, B).

FIGURE 4-25. A, Reflective markers are used to indicate anatomic locations for determination of joint angular displacement of a walking individual. Marker location is acquired using video-based cameras that can operate at variable sampling rates. B, A computerized animated “stick figure” generated by data collected from the subject shown in A. (Courtesy Vicon Motion Systems, Inc., Centennial, Colorado.)

Video-based systems are quite versatile and are used to analyze human functional activities ranging from whole-body motion (e.g., swimming, running) to smaller motor tasks (e.g., typing, reaching). Some systems allow movement to be captured outdoors and processed at a later time, whereas others can process the signal almost in real time. Another desirable feature of most video-based systems is that the subject is not encumbered by wires or other electronic devices.

Optoelectronics is another popular type of kinematic acquisition system that uses active markers that are pulsed sequentially. The light is detected by special cameras that focus it on a semiconductor diode surface. The system enables collection of data at high sampling rates and can acquire real-time, 3D data. The system is limited in its ability to acquire data outside a controlled environment. Subjects may feel hampered by the wires that are connected to the active markers. Telemetry systems enable data to be gathered without the subjects being tethered to a power source, but these systems are vulnerable to ambient electrical interference.

Electromagnetic Tracking Devices: Electromagnetic tracking devices measure six degrees of freedom (three rotational and three translational), providing position and orientation data during both static and dynamic activities. Small sensors are secured to the skin overlying anatomic landmarks. Position and orientation data from the sensors located within a specified operating range of the transmitter are sent to the data capture system.

One disadvantage of this system is that the transmitters and receivers can be sensitive to metal in their vicinity that distorts the electromagnetic field generated by the transmitters. Although telemetry is available for these systems, most operate with wires that connect the sensors to the data capture system. The wires limit the volume of space from which motion can be recorded.

In any motion analysis system that uses skin sensors to record underlying bony movement, there is the potential for error associated with the extraneous movement of skin and soft tissue.

KINETIC MEASUREMENT SYSTEMS

Mechanical Devices: Mechanical devices measure an applied force by the amount of strain of a deformable material. Through purely mechanical means, the strain in the material causes the movement of a dial. The numeric values associated with the dial are calibrated to a known force. Some of the most common mechanical devices for measuring force include a bathroom scale, a grip strength dynamometer, and a hand-held dynamometer (as shown in Figure 4-19).

Transducers: Various types of transducers have been developed and widely used to measure force. Among these are strain gauges and piezoelectric, piezoresistive, and capacitance transducers. Essentially these transducers operate on the principle that an applied force deforms the transducer, resulting in a change in voltage in a known manner. Output from the transducer is converted to meaningful measures through a calibration process.

One of the most common transducers for collecting kinetic data while a subject is walking, stepping, or running is the force plate. Force plates use piezoelectric quartz or strain-gauge transducers that are sensitive to load in three orthogonal directions (an example of a force plate is shown in Figure 4-27, ahead, under the subject’s forward right foot). The force plate measures the ground reaction forces in vertical, medial-lateral, and anterior-posterior components. The ground reaction force data are used in subsequent dynamic analysis.

Electromechanical Devices: A common electromechanical device used for dynamic strength assessment is the isokinetic dynamometer. During isokinetic testing, the device maintains a constant angular velocity of the tested limb while measuring the external torque applied to resist the subject’s produced internal torque. The isokinetic system can often be adjusted to measure the torque produced by most major muscle groups of the body. Most isokinetic dynamometers can measure kinetic data produced by concentric, isometric, and eccentric activation of muscles. The angular velocity is determined by the user, varying between 0 degrees/sec (isometric) and 500 degrees/sec during concentric activations. Figure 4-26 shows a person who is exerting maximal-effort knee extension torque through a concentric contraction of the right knee extensor musculature. Isokinetic dynamometry provides an objective record of muscular kinetic data, produced during different types of muscle activation at multiple test velocities. The system also provides immediate feedback of kinetic data, which may serve as a source of biofeedback during training or rehabilitation.

FIGURE 4-26. Isokinetic dynamometry. The subject is generating a maximal-effort knee extension torque at a joint angular velocity of 60 degrees/sec. The machine is functioning in its “concentric mode,” providing resistance against the contracting right knee quadriceps muscle. Note that the medial-lateral axis of rotation of the tested knee is aligned with the axis of rotation of the dynamometer. (Courtesy of Biodex Medical Systems, Shirley, New York.)

SPECIAL FOCUS 4-7 Introduction to the Inverse Dynamic Approach for Solving for Internal Forces and Torques

Measuring joint reaction forces and muscle-produced net torques during dynamic conditions is often performed indirectly using a technique called the inverse dynamic approach.²² A direct dynamic approach determines accelerations and external forces and torques through knowledge of internal forces and torques. Conversely, an inverse dynamic approach determines internal forces and torques through knowledge of accelerations and external forces and torques. The inverse dynamic approach relies on data from anthropometry, kinematics, and external forces, such as gravity and contact forces. Accelerations are determined employing the first and second derivatives of position-time data to yield velocity-time and acceleration-time data, respectively. The importance of acquiring accurate position data is a prerequisite to the soundness of this approach, because errors in measuring position data magnify errors in velocity and acceleration.

In the inverse dynamics approach, the system under consideration is often defined as a series of link segments. Figure 4-27, A illustrates the experimental setup for investigating the forces and torques in the right lower limb during different versions of a forward lunge exercise with three different trunk and upper extremity positions.⁵ To simplify calculations, the subject’s right lower limb is considered a linked segment model consisting of solid foot, leg, and thigh segments linked by frictionless hinges at the ankle and knee, and to the body at the hip (see Figure 4-27, B). The center of mass (CM) is located for each segment. In Figure 4-27, C the modeled segments of the right lower limb are disarticulated and the individual forces and torques (moments) are identified at each segment end point. The analysis on the series of links usually begins with the analysis of the most distal segment, in this case the foot. Information gathered through motion analysis techniques, typically camera based, serves as input data for the dynamic equations of motion (Equations 4-15 and 4-16). This information includes the position and orientation of the segments in space and the acceleration of the segments and the segments’ centers of mass. The ground reaction forces (components G_Y and G_X) acting on the distal end of the segment are measured in this example by a force plate built into the floor. From these data the ankle joint reaction force (components JA_Y and JA_X) and the net muscle torque (moments) at the ankle joint are determined. This information is then used as input for continued analysis of the next most proximal segment, the leg. Analysis takes place until all segments or links in the model are studied. Several assumptions made during the use of the inverse dynamic approach are included in the box.

FIGURE 4-27. Example of an inverse dynamic approach to kinetic analysis of three versions of a forward lunge. A, Photograph of the experimental setup with the subject lunging onto the force plate with her right leg. Images were superimposed to show the three different trunk and upper extremity positions of interest. Videography-based passive reflective markers used to collect motion analysis data are visible on the lateral aspect of the subject’s right shoe and on cuffs attached to her leg and thigh. Wires are also visible connecting electromyographic electrodes overlying the subject’s muscles to a telemetry unit worn on the subject’s back. B, The link model of the lower limb is shown as consisting of three articulated segments: thigh (T), leg (L), and foot (F). The center of mass (CM) of each segment is represented as a fixed point (red circle): CM_T, CM_L, and CM_F. C, The three link segments are disarticulated in order for the internal forces and torques to be determined, beginning with the most distal foot segment. The red curved arrows represent torque (moment) around each axis of rotation: M_A, M_K, and M_H are moments at the ankle, knee, and hip respectively; W_F, W_L, and W_T are segment weights of foot, leg, and thigh, respectively; JA_X and JA_Y, JK_X and JK_Y, and JH_X and JH_Y are joint reaction forces at the ankle, knee, and hip, respectively; G_X and G_Y are ground reaction forces acting on the foot. The coordinate system is set up with X horizontal and Y vertical; arrowheads point in positive directions. (A from Farrokhi S, Pollard C, Souza R, et al: Trunk position influences the kinematics, kinetics, and muscle activity of the lead lower extremity during the forward lunge exercise, J Orthop Sports Phys Ther 38:403, 2008.)

Assumptions Made during the Inverse Dynamic Approach

1. Each segment or link has a fixed mass that is concentrated at its center of mass.

2. The location of each segment’s center of mass remains fixed during the movement.

3. The joints in this model are considered frictionless hinge joints.

4. The mass moment of inertia of each segment is constant during the movement.

5. The length of each segment remains constant.

SUMMARY

Many evaluation and treatment techniques used in rehabilitation involve the application or generation of forces and torques. A better understanding of the rationale and consequences of these techniques can be gained through the application of Newton’s laws of motion and through static equilibrium or dynamic analyses. Although it is recognized that formal analyses are rarely completed in a clinic setting, principles learned from these analyses are clinically important and applied often. For example:

• Changing the moment of inertia of an arm by bending or straightening the elbow changes the required torque to move the shoulder.

• During an exercise the forces generated by muscles are often many times greater than the external forces used as resistance. This must be considered when a damaged muscle or tendon is being exercised.

• External torque is minimal when the line of force of the external force passes through or near the axis of motion.

• External torque is maximal when the line of force of the external force is at right angles to the limb. When gravity is used as a resistance force, this occurs when the limb is in a horizontal position.

• Internal torque produced by a muscle is maximal when its angle-of-insertion is 90 degrees.

• Exercises are often optimized when external and internal torques are matched through a range of motion.

• Forces at a joint occur as a necessary reaction to the combination of internal and external forces. Muscle force often plays the dominant role in the creation of these joint reaction forces.

Three quantitative-based sample problems were highlighted in this chapter. Two additional problems are available in Appendix I, Part B.

Additional Clinical Connections

CLINICAL CONNECTION 4-1 A Practical Method for Estimating Relative Torque Potential Based on Leverage

Earlier in this chapter, Figures 4-15 and 4-16 showed two methods for estimating internal and external torques. In both figures, Method 2 is considered a “shortcut” method because the resolution of the resultant forces into their component forces is unnecessary. Consider first internal torque (see Figure 4-15). The internal moment arm (depicted as IMA_M)—or leverage—of most muscles in the body can be qualitatively assessed by simply visualizing the shortest distance between a given whole muscle’s line of force and the associated joint’s axis of rotation. This experience can be practiced with the aid of a skeletal model and a piece of string that represents the resultant muscle’s line of force (Figure 4-28). As apparent in the figure, the internal moment arm (shown in brown) is greater in position A than in position B; this means that for the same biceps force, more internal torque will be generated in position A than in position B. In general, the internal moment arm of any muscle is greatest when the angle-of-insertion of the muscle is 90 degrees to the bone.

FIGURE 4-28. A piece of string can be used to mimic the line of force of the resultant force vector of an activated biceps muscle. The internal moment arm is shown as a brown line; the axis of rotation at the elbow is shown as a solid black circle. Note that the moment arm is greater when the elbow is in position A compared with position B. (Modified from LeVeau BF: Williams & Lissner’s biomechanics of human motion, ed 3, Philadelphia, 1992, Saunders.)

Next, consider the shortcut method for determining external torque. Clinically, it is often necessary to quickly compare the relative external torque generated by gravity or other external forces applied to a joint. Consider, for example, the external torque at the knee during two squat postures (Figure 4-29). By visualizing the external moment arm between the knee joint axis of rotation and the line of gravity from body weight, it can be readily concluded that the external torque is greater in a deep squat (A) compared with a partial squat (B). The ability to judge the relative demand placed on the muscles because of the external torque is useful in terms of protecting a joint that is painful or otherwise abnormal. For instance, a person with arthritic pain between the patella and femur is often advised to limit activities that involve lowering and rising from a deep squat position. This activity places large demands on the quadriceps muscle, which increases the compressive forces on the joint surfaces.

FIGURE 4-29. The depth of a squat significantly affects the magnitude of the external torque at the knee produced by superimposed body weight. The relative external torque, within the sagittal plane, can be estimated by comparing the distance that the body weight force vector falls posteriorly with the medial-lateral axis of rotation at the knee (shown as open circle). The external moment arm (EMA)–and thus the external torque created by body weight–is greater in A than in B. The external moment arms are shown as brown lines, originating at the axis of rotation and intersecting body weight force at right angles.

CLINICAL CONNECTION 4-2 Modifying Internal Torque as a Means to Provide “Joint Protection”

Some treatments in rehabilitation medicine are directed at reducing the magnitude of force on joint surfaces during the performance of a physical activity. The purpose of such treatment is to protect a weakened or painful joint from large and potentially damaging forces. This result can be achieved by reducing the rate of movement (power), providing shock absorption (e.g., cushioned footwear), or limiting the mechanical force demands on the muscle.

Minimizing large muscular-based joint forces may be important for persons with a prosthesis (artificial joint replacement). A person with a hip replacement, for example, is often advised on ways to minimize unnecessarily large forces produced by the hip abductor muscles.12,14,15 Figure 4-30 depicts a simple schematic representation of the pelvis and femur while a person with a prosthetic hip is in the single-limb support phase of gait. In order for equilibrium to be maintained within the frontal plane, the internal (counterclockwise, +Z) and the external (clockwise, −Z) torques around the stance hip must be balanced. As shown in both the anatomic (A) and see-saw (B) illustrations of Figure 4-30, the product of the body weight (W) times its moment arm D₁, must be equal in magnitude and opposite in direction to the hip abductor muscle force (M) times its moment arm (D): W × D₁ = M × D. Note that the external moment arm around the hip is almost twice the length of the internal moment arm.^16,17 This disparity in moment arm lengths requires that the muscle force be almost twice the force of superincumbent body weight in order to maintain equilibrium. In theory, reducing excessive body weight, carrying lighter loads, or carrying loads in certain fashions can decrease the external force and/or the external moment arm and therefore decrease the external torque about the hip. Reduction of large external torques substantially decreases large force demands from the hip abductors, thereby decreasing joint reaction forces on the prosthetic hip joint.

FIGURE 4-30. A, Hip abductor muscle force (M) produces a torque necessary for the frontal plane stability of the pelvis during the right single-limb support phase of gait. Rotary stability is established, assuming static equilibrium, when the external (clockwise, −Z) torque created by superincumbent body weight (W) is exactly balanced by the internal (counterclockwise, +Z) torque from the hip abductor muscles (M). The counterclockwise torque equals M times its moment arm (D), and the clockwise torque equals W times its moment arm (D₁). B, This first-class lever seesaw model simplifies the model shown in A. The joint reaction force (J), assuming that all force vectors act vertically, is shown as an upward-directed force with a magnitude equal to the sum of the hip abductor force and superimposed body weight. The X-Y coordinate reference frame is placed so the X axis is parallel with the body weight (W); thin black arrowheads point toward the positive direction. (Modified from Neumann DA: Biomechanical analysis of selected principles of hip joint protection, Arthritis Care Res 2:146, 1989.)

Certain orthopedic procedures illustrate how concepts of joint protection are used in rehabilitation practice. Consider the case of severe hip osteoarthritis, which results in destruction of the femoral head and a subsequent reduced size of the femoral neck and head. The bony loss shortens the internal moment arm length (D in Figure 4-30, A) available to the hip abductor muscles (M); thus, greater muscle forces are needed to maintain frontal plane equilibrium, and greater joint reaction forces result. A surgical procedure that attempts to reduce joint forces on the hip entails the relocation of the greater trochanter to a more lateral position. This procedure increases the length of the internal moment arm of the hip abductor muscles. An increase in the internal moment arm reduces the force required by the abductor muscles to generate a given torque during the single-limb support phase of gait.

CLINICAL CONNECTION 4-3 The Influence of Antagonist Muscle Coactivation on the Clinical Measurement of Torque

When muscle strength is measured, care must be taken to encourage activation of the agonist muscles and relative relaxation of the antagonist muscles (review definitions of agonist and antagonist muscles in Chapter 1). Coactivation of antagonist muscles alters the net internal torque and decreases the ability to control or overcome external forces and torques. This concept is shown with the use of a hand-held dynamometer, similar to that previously described in Figure 4-19. Figure 4-31, A shows the measurement of elbow extension torque with activation only from the agonist (elbow extensor) muscles while the antagonist elbow flexors are relaxed. In contrast, Figure 4-31, B show a maximal-effort strength test of the elbow extensors with coactivation of both the agonist (E) and antagonist elbow flexor (F) muscles. (This situation may occur in a healthy person who is simply unable to relax the antagonist muscles or in a patient with neurologic pathology such as Parkinson’s disease or cerebral palsy.) The internal torque produced by the antagonist muscles actually subtracts from the internal torque produced by the agonist muscles. As a result, the net internal torque is reduced, as indicated by the reduced external force (R) applied against the dynamometer. Because the test condition is isometric, the measured external torque is equal in magnitude but opposite in direction to the reduced net internal torque. The important clinical point here is that even though the elbow extensor forces and torques may be equivalent in tests A and B of Figure 4-31, the external torque measures less in B. This scenario may give an erroneous impression of relative weakness of the agonist muscles when, in fact, they are not weak. As always, the joint reaction force (J) occurs in response to the sum of all forces across the joint and therefore will be increased in test B with antagonist activation.

FIGURE 4-31. The influence of coactivation of the agonist (elbow extensor) and antagonist (elbow flexor) muscle groups is shown on the apparent strength (torque) of isometric elbow extension. A, Agonist (elbow extensor) activation only, with the same conditions and abbreviations used in Figure 4-19, A. B, Subject is simultaneously coactivating her elbow extensors and (antagonistic) elbow flexors muscles, producing a simultaneous elbow extension force (E) and an elbow flexion force (F). Because F and E generate oppositely directed torques around the elbow, the net elbow extension torque is reduced. Note, however, that the magnitude of the joint reaction force (J) is increased in B. Vectors are drawn to approximate scale. Based on conventions summarized in Box 4-1, the internal moment arm used by the extensor muscles is assigned a negative number. This, in turn, assigns opposite rotational directions to the opposing internal torques. EMA, external moment arm; IMA_F and IMA_E, internal moment arms of the elbow flexors and extensor muscles, respectively; R, external force measured by the dynamometer.