Joint Structure & Function-A Comprehensive Analysis Fourth Edition Pamela K.

Copyright © 2005 by F. A. Davis. 30 ■ Section 1: Joint Structure and Function: Foundational Concepts TF (166 N) site acceleration of segments) or static (when the tensile forces in the tissues that join the segments are balanced by distraction forces of equal or greater magnitude). GF (78 N) CF (88 N) ■ Joint Compression and Joint Reaction Forces GWbLf (88 N) Supporting Sam Alexander’s leg-foot segment can ▲ Figure 1-41 ■ Distraction of the joint and tensile forces in minimize or eliminate the tension in his injured capsu- the knee joint capsule occurs when there is a net distractive force loligamentous structures. In Figure 1-42, the support- directed away from the joint surfaces applied to each of the adjacent ing upward push of the hand on the leg-foot segment joint segments (dashed vectors). The distractive force on the femur is has been increased to ϩ90 N. Given that the magnitude provided by the force of table-on-femur (TF), whereas the distractive of GWbLf (resultant of gravity and weight boot) is still force on the leg-foot segment is provided by GWbLf. Ϫ88 N, these two forces will result in a net unbalanced force on the leg-foot segment of ϩ2 N. The leg-foot should be the sum of the magnitudes of CF and GF segment will accelerate upward until a new force is en- (but opposite in direction). However, vectors CF, GF, countered. This new force cannot come from the cap- and TF are not in a linear force system because they sule that is now becoming increasingly slack, but it will are not co-linear. Rather, they are parallel forces. arise once the leg-foot segment makes contact with the Although we will tackle composition (and the femur. The upward acceleration of the leg-foot segment effects) of parallel forces in more detail later, we can will stop when the contact force, femur-on-legfoot use the same shorthand system here that we used to (FLf), reaches a magnitude of Ϫ2 N (see Fig. 1-42), at compose two gravitational vectors earlier in the which point equilibrium of the leg-foot segment is chapter. restored. Because both GF and CF (see Fig. 1-41) are ver- When the two segments of a joint are pushed tically downward, the resultant of these two forces together and “touch,” as occurs with the upward sup- would be a new downward force with the combined port of the hand in Figure 1-42 (legfoot-on-femur and magnitudes of the original two (88 N ϩ 78 N), with femur-on-legfoot), the resulting reaction (contact) a point of application along a line drawn between forces are also referred to as joint reaction forces.3 Joint the original two points and located slightly toward reaction forces are contact forces that result whenever the vector with the greater magnitude. Because this two or more forces cause contact between contiguous new resultant vector will lie approximately in line joint surfaces. Joint reaction forces are dependent on with vector TF, we now have two forces in a linear the existence of one force on each of the adjacent joint force system on an object in equilibrium. Therefore, segments that is perpendicular to and directed toward vector TF must have a magnitude of ϩ166 N. Vector its joint surface. The two forces that cause joint reac- TF must be the second distraction force because it is tions forces are known as compression forces. Com- applied perpendicular to and away from the joint pression forces are required to push joint surfaces surface. In Figure 1-41, the two distraction forces together to produce joint reaction forces in the same (GWbLf and TF) are shown as dashed vectors. way that distraction forces are required to produce cap- suloligamentous or muscular tension across separating CONCEPT CORNERSTONE 1-8: Joint Distraction and (or separated) joint surfaces. [Side-bar: It is important Distraction Forces to note that the term “compression” here refers to pushing together rigid nondeformable bones to close a ■ Distraction forces create separation of joint surfaces. joint space. Compression across or within a deformable ■ There must be a minimum of one (or one resultant) distrac- body is more complex and will be considered in Chapter 2.] tion force on each joint segment, with each distraction force perpendicular to the joint surfaces, opposite in direction to In Figure 1-42, one of the forces causing joint com- the distraction force on the adjacent segment, and directed pression at the knee joint is hand-on-legfoot (HLf) away from its joint surface. because HLf is applied toward the articulating surface ■ Joint distraction can be dynamic (through unequal or oppo- of the leg-foot segment and is perpendicular to that sur- face. If, however, the ϩ2 N push of the leg-foot segment on the femur is not offset by a downward force of at least 2 N on the femur, the femur will also accelerate upward. If the femur and leg-foot segment were to accelerate upward at the same rate (and in the same direction), the contact between the joint surfaces might be main- tained but could not be greater than 2 N. [Side-bar: Although the leg-foot segment is our focus, rather than the femur, it is worth noting that the femur is not likely to move because gravity is acting downward on the femur to stabilize it with a force of 78 N (see Fig. 1-42). Gravity-on-femur is the second joint compression force because it is the only force on the femur that is applied

Copyright © 2005 by F. A. Davis. Chapter 1: Biomechanical Applications to Joint Structure and Function ■ 31 LFf (2 N) ■ There must be a minimum of one (or one resultant) com- pression force on each contiguous joint segment, with each compression force perpendicular to and directed toward the segment’s joint surface, and opposite in direction to the compression force on the adjacent segment. FLf (2 N) Revisiting Newton’s Law of Inertia GF (78 N) It would appear that the weight boot is a poor option for Sam, given the potential tensile forces created in his GWbLf (88 N) injured joint capsule (and ligaments), unless we plan to continue supporting his leg-foot segment with a hand HLf (90 N) (or, perhaps, a bench). However, it has been assumed thus far that Sam is relaxed. As soon as Sam initiates a ▲ Figure 1-42 ■ Joint compression results in joint reaction contraction of his quadriceps muscle, the balance of forces (FLf and LfF) when there is a net compression force applied forces will change. Before we add the muscle force to to each of the adjacent joint segments (dashed vectors) toward the the weight boot exercise, however, let us return to the joint surfaces, in this case provided by hand-on-legfoot (HLf) and leg-press exercise to identify what effect, if any, the gravity-on-femur (GF). forces from the leg press will have on the leg-foot segment or Sam’s injured capsuloligamentous struc- perpendicular to and toward the joint surface. In tures. Figure 1-42, the two joint compression forces are shown as dashed vectors.] In the leg-press exercise, Sam Alexander’s leg-foot segment is contacting the footplate of the leg-press Whenever there is a net compression of joint sur- machine, creating the force of footplate-on-legfoot faces (resulting in joint reaction forces), the capsule (FpLf). The magnitude of vector FpLf is not yet known. and ligaments at the joint are generally not under ten- There are also other forces acting on the leg-foot seg- sion (as long as all forces are perpendicular to contact- ment because other things are touching the leg-foot ing surfaces). The pull of capsule-on-legfoot segment is segment. One of these is gravity. Two other options are not shown in Figure 1-42 because the tension in the contacts of femur-on-legfoot or capsule-on-legfoot. capsule has effectively been eliminated (or reduced to Whether the push of the femur on the leg-foot segment imperceptible magnitude). Equilibrium between two or the pull of the capsule on the leg-foot segment is a bony segments with net joint compression and equal factor in this space diagram requires further explo- and opposite joint reaction forces also assumes that the ration. We will begin with the known force, gravity-on- push of one bony segment on another does not result legfoot (GLf). in failure of the bone (that is, that one bone does not accelerate through the other). The magnitude of the weight of the leg-foot seg- ment remains the same as in the weight boot example Continuing Exploration: Close-Packing of a Joint (Ϫ88 N), but the orientation to the leg-foot segment differs. Consequently, the orientation of gravity to the Although capsuloligamentous structures are typi- leg-foot segment differs (Fig. 1-43). The force of foot- cally not under tension when there are net compres- plate-on-legfoot (FpLf) is also shown in the figure but sive forces (with no shear forces) across a joint, there has no designated magnitude because the magnitude is is an important exception. With sufficient twisting of not yet unknown. Vectors GLf and FpLf cannot be the capsuloligamentous structures of a joint, the summed to find their resultant effect because the two adjacent articular surfaces are drawn into contact by the pull of the capsule on the bony segments. This is FY FH called “close-packing” of the joint. This concept will be elaborated upon in Chapter 2 and in examination FpLf of the individual joint complexes. CONCEPT CORNERSTONE 1-9: Joint Compression and GLf (48 N) Joint Compression Forces ▲ Figure 1-43 ■ The known forces of footplate-on-legfoot ■ Joint compression forces create contact between joint sur- (FpLf) and gravity-on-legfoot (GLf) must be balanced by another faces. horizontal (FH) and vertical (FV) force, respectively.

Copyright © 2005 by F. A. Davis. 32 ■ Section 1: Joint Structure and Function: Foundational Concepts forces are not in the same linear force system (they are rigid (nondeformable) structures (e.g., bones). Shear not co-linear). It is theoretically possible to find the within deformable structures will be considered in resultant of these two forces through composition by Chapter 2.] parallelogram because these two vectors are part of a concurrent force system (the vectors will intersect if the A friction force (Fr) potentially exists on an object vectors are extended). That solution, however, requires whenever there is a contact force on that object. that we know at least the relative magnitudes of FpLf Friction forces are always parallel to contacting surfaces and GLf. A second option is to consider the two differ- (or tangential to curved surfaces) and have a direc- ent linear force systems of which vectors GLf and FpLf tion that is opposite to potential movement. For fric- are a part and determine the magnitudes of the vectors tion to have magnitude, some other force (a shear within each linear force system. force) must be moving or attempting to move one or both of the contacting objects on each other. The force ■ Vertical and Horizontal Linear Force Systems of friction can be considered a special case of a shear force because both are forces parallel to contacting Newton’s law of inertia (or law of equilibrium) can be surfaces, but friction is a shear force that is always in broken down into component parts: The sum of the the direction opposite to movement or potential vertical forces (FV) acting on an object in equilibrium movement. must total zero (∑FV ϭ 0), and, independently, the sum of the horizontal forces (FH) acting on an object in Whenever a shear force (FS1) is present on an equilibrium must total zero (∑FH ϭ 0). Consequently, object, there will always be at least one opposing shear there must be at least two additional forces acting on force (FS2) on that object. In the absence of an op- the leg-foot segment that are equal in magnitude and posing shear force created by the contact of a new opposite in direction to GLf and FpLf because the leg- object, the opposing shear force (FS2) will be friction foot segment cannot be at rest unless the sum of forces (Fr). If the magnitude of an opposing shear force (FS2) in both linear forces systems equals zero. Forces FV and created by a contact of a new object is inadequate FH are drawn in Figure 1-43, but the source of each to prevent movement by FS1, friction (FS3) will also force is not yet established. oppose FS1. We know that the femur and the capsule are both Sam Alexander’s leg-foot segment is contacting the contacting and potentially creating forces on the leg- footplate. Because FpLf is a contact (or normal) force, foot segment. Given these options, it appears that vec- it can also be labeled FC (Fig. 1-44). The force of tor FH is likely to be the push of the femur on the gravity-on-legfoot is parallel to the foot and footplate leg-foot segment (FLf) because the pull of capsule-on- and has the potential to slide the foot down the foot- legfoot would be in the opposite direction. The magni- plate. Consequently, gravity-on-legfoot (GLf) may also tude of FLf and FpLf can be estimated to be fairly small be referred to as a shear force. In the absence of if Sam is relaxed and the footplate is locked in position. another other opposing shear, there will be a concomi- Before attempting to determine the magnitude of FLf tant opposing force of friction-on-legfoot (FrLf) that is and FpLf, we will examine the source and magnitude of parallel to the foot and footplate surfaces and is in a FV because it will be seen that, in this example, FV and direction opposite to the potential slide of the foot (see FH are related. Fig. 1-44). To understand the magnitude of FrLf, we need to further explore the force of friction. The source of the FV is difficult to ascertain because it appears that we have accounted for all objects con- FrLf tacting the leg-foot segment, including gravity, with FLf none that appear to act in the direction of FV. To iden- tify FV, we must acknowledge an additional property of FrFp all contact forces. Whenever there is contact between two objects (or surfaces of objects), the potential exists FpLf (FC) for friction forces on both contacting surfaces. The fric- tion forces will have magnitude, however, only if there are concomitant opposing shear forces on the contact- ing objects. Shear and Friction Forces GLf (FS) (48 N) A force (regardless of its source) that moves or attempts ▲ Figure 1-44 ■ Footplate-on-legfoot (FpLf) is a contact force to move one object on another is known as a shear (FC) that will result in friction-on-legfoot (FrLf) between the foot and force (FS). A shear force is any force (or the compo- footplate, given the shear force (FS), GLf. Femur-on-legfoot (FLf) is nent of a force) that is parallel to contacting surfaces also a contact force, but the low coefficient of friction for articular (or tangential to curved surfaces) and has an action cartilage makes the value of friction between the femur and leg-foot line in the direction of attempted movement. [Side-bar: segment negligible. Shown in a shaded vector that is not part of the The discussion here is on shear forces between two space diagram is the reaction force to FrLf, friction-on-footplate (FrFp).

Copyright © 2005 by F. A. Davis. Chapter 1: Biomechanical Applications to Joint Structure and Function ■ 33 ■ Static Friction and Kinetic Friction remains unchanged because the surface remains skin on skin). [Side-bar: It is commonly thought that the The magnitude of a friction force on an object is always magnitude of friction between two surfaces is related to a function of the magnitude of contact between the the amount of surface area in contact (pressure or objects and the slipperiness or roughness of the con- force per unit area). However, the only contributing tacting surfaces. When two contacting objects with factors are the magnitude of contact and the coefficient shear forces applied to each are not moving, the magni- of the contacting surfaces. The area of contact does not tude of friction on each object is also proportional to affect the magnitude of friction.2] the magnitude of the shear forces. If the two objects are not moving (objects are static), the maximum magnitude In Figure 1-45A, a large box weighing 445 N (~45 of the force of static friction (Frs) on each object is the kg or 100 lb) is resting on the floor. The floor must product of a constant value known as the coefficient of push on the box (FB) with a magnitude equal to the static friction (␮S) and the magnitude of the contact weight of the box (GB) because the box is not moving force (FC) on each object; that is, (∑FV ϭ 0). Because nothing is attempting to move the box parallel to the contacting surfaces (bottom of the FrS р ␮SFC box and the floor), there will be no friction on either the box or the floor. However, as soon as the man The coefficient of static friction is a constant value begins to push on the box (see Fig. 1-45B), the man’s ifasoprapsrgoilvixtetimnleamatsealt0ye.2r0i5.a0.l55s.;AtFshoetrhveeaxlcuaomenpotaflce␮t,inS␮gfSosrfuowrrfoaioccedesoobnnecwicooemoides force (man-on-box [MB]) creates a shear force, with a softer or rougher, ␮S increases. As the magnitude of concomitant resulting force of friction-on-box (FrB). contact (FC) between objects increases, so too does the (Note that Fig. 1-45B is oversimplified because the FrB magnitude of potential friction. The greater the con- is shown acting in line with MB, rather than at the bot- tact force on an object is and the rougher the contact- tom of the box as would actually be the case.) Assuming ing surfaces are, the greater the maximum potential that the man’s initial push is not sufficient to move the force of friction is. When using friction to warm your box, we can begin by calculating the maximum possible hands, the contact of the hands warms both of them magnitude of friction-on-box. (friction forces exist on both the right and the left hands). If you wish to increase the friction, you press The maximum friction force on the box when the your hands together harder (increase the contact box is not moving is a product of the coefficient of force) as you rub. Increasing the pressure increases the static friction of wooden box on wood floor (0.25) and contact force between the hands and increases the max- the magnitude (445 N) of the contact of floor-on- imum value of friction (the coefficient of friction box (FB): FrB р (0.25)(445 N) FrB р 111.25 N FB (445 N) FB (445 N) GB (-445 N) MB FrB GB (-445 N) AB ▲ Figure 1-45 ■ A. The box is acted on by the forces of gravity (GB) and the contact of floor-on-box (FB). The force of friction has no magnitude (so is not shown) because there is no attempted movement. B. The force of the man-on-box (MB) causes an opposing friction force (FrB).