The equilibrium established in John Alexander’s leg-foot segment as he sits with the dangling weight boot is de- pendent on the capsule (and ligaments) to pull upward on the leg-foot segment with the same magnitude as that with which gravity and the weight boot pull downward (see Fig. 1–26C). Because the capsuloligamentous struc- tures are injured in John’s case, we need to explore the forces applied to the capsule. If the capsule pulls on the leg-foot segment with a magnitude of 88 N, the law of reaction stipulates that the leg-foot segment must also be pulling on the capsule with an equivalent force. If the compromised capsule cannot withstand an 88-N pull, then it may not be able to pull on the leg-foot segment with an 88-N force. This relationship requires an under- standing of tensile forces and the forces that produce them.
Tensile Forces
Tension in the joint capsule, just like tension in any passive structure (including relatively solid materials such as bone), is created by opposite pulls on the same object. If there are not two opposite pulls on the object (each of which is a tensile force), there cannot be tension in the object. Remembering that the connective tissue capsule and ligaments are best analogized to slightly elasticized cord, we first examine tension in a cord or rope.
two forces can be established by using both Newton’s first and third laws.
The person standing on the scale must be in equilib- rium (ΣF = 0). If the gravity-on-person vector is acting down with a magnitude of –734 N (John Alexander’s weight), there must also be a force of equal magnitude acting up on the person for the person to remain motion- less. The only other object, besides gravity, that appears to be contacting the person in Figure 1–29 is the scale.
The scale, therefore, must be exerting an upward push on the person (scale-on-person [SP]) with magnitude equal to that of gravity-on-person (+734 N). The force of scale- on-person, of course, has a reaction force of person-on- scale that is equal in magnitude (734 N) and opposite in direction (down) but applied to the scale. Consequently, in this instance, the magnitude of the person’s weight and the magnitude of the person’s contact with the scale are equivalent although applied to different objects. The vectors person-on-scale and scale-on-person occur as a result of a push by the contacting objects. When reaction forces arise from the push of one object on another, they are of- ten referred to as contact forces (FC). When contact forces are perpendicular to the surfaces that produce them, the term normal force (FN) is also used.5,9 Contact forces, therefore, are a subset of reaction forces.
The distinction between gravity-on-person and the re- action force person-on-scale is not always made, but it can be very important if something else is touching the person or the scale. If the person is holding something while on the scale, the person’s weight does not change, but the contact forces between the person and the scale will in- crease. Similarly, a gentle pressure down on the bathroom countertop as a person stands on the scale will result in an apparent weight reduction. Situations are frequently encountered in which the contact of an object with a sup- porting surface and its weight are used interchangeably.
Figure 1–29 Although a scale is commonly thought to measure the weight of the person (gravity-on-person [GP]), it actually measures the contact of the person-on-scale (PS). Vectors GP and PS are equal in magnitude as long as nothing else is touching the person.
ConceptCornerstone 1-6
Action-Reaction Forces
• Whenever two objects or segments touch, the two objects or segments exert a force on each other. Con- sequently, every force has a reaction or is part of an action-reaction pair.
• The term contact force or contact forces is commonly used to indicate one or both of a set of reaction forces in which the “touch” is a push rather than a pull.
• Reaction forces are never part of the same force system and cannot be composed (cannot either be additive or offset each other) because the two forces are, by definition, applied to different objects.
• The static or dynamic state (equilibrium or motion) of an object cannot be affected by another object that is not touching it or by a force that is not applied to it.
• The reaction to a force should be acknowledged but may be ignored graphically and conceptually if the object to which it is applied and the other forces on that object are not of interest.
Care should be taken to assess the situation to determine whether the magnitudes are, in fact, equivalent. The recognition of weight and contact as separate forces per- mits more flexibility in understanding how to modify these forces if necessary.
If a man pulls on a rope that is not attached to any- thing, no tension will develop in the rope no matter how hard or lightly he pulls because there is no counterforce.
The rope will simply accelerate in the direction of the man’s pull (with a magnitude equivalent to the force of pull [Funbal] divided by the mass [m] of the rope). If the rope is tied to an immovable block of cement, there will be two forces applied to the rope. The two forces are created by the only two things contacting the rope: the man’s hands and the block (Fig. 1–30). If the hands-on- rope vector (HR) has a magnitude of +110 N (~25 lb), then the block-on-rope vector (BR) must have a magni- tude of –110 N because the rope is in equilibrium. If it is assumed that rope has a homogenous composition (unlike most biological tissues), the tension will be the same throughout the rope (as long as there is no friction on the rope), and the tension in the rope will be equivalent to the magnitude of the two tensile forces acting on the rope.5In
Figure 1–30, both hands-on-rope (+110 N) and block-on- rope (–110 N) can be designated as tensile forces.
Assume for the moment that the rope is slack before the man begins to pull on the rope. As the man initiates his pull, his hands will accelerate away from the block because the force pulling his hands toward his body (muscles-on-hands [MsH]) will be greater than the pull of the rope on the hands (rope-on-hands [RH]). As the man’s hands get farther from the block, the rope will get tighter, and the force of the rope-on-hands will increase. The acceleration of the hands will gradually slow down as the resultant of these two forces approaches equilibrium;
when the two forces are equal, there will be zero acceler- ation. Because rope-on-hands and hands-on-rope are re- action forces (and always equal in magnitude), the tension in the rope (hands-on-rope) will eventually be equivalent to the magnitude of the man’s pull (muscles-on-hands) (Fig. 1–31).
Tensile Forces and Their Reaction Forces
The example of the man pulling on the rope and cement block assumed that the rope could withstand whatever ten- sion was required of it. If the rope is damaged, it may be able to withstand no more than the required 110 N of tension. If, however, the man in our example pulls on the rope with a magnitude of 200 N (45 lb), the rope will break. Once the rope breaks, there is no longer tension in the rope. The man pulling on the rope with a magnitude of 200 N will have a net unbalanced force that will accelerate his hands (or the man) backward (Fig. 1–32) until his muscles stop pulling (which will, it is hoped, happen before he punches himself in the stomach or falls over!).
Let us go back to John Alexander to determine how the tension example is applied to John’s use of the weight boot. The equilibrium of John’s leg-foot segment was based on the ability of the capsule to pull on the leg-foot segment with a magnitude equivalent to gravity-on-leg- foot plus weightboot-on-legfoot. If the capsule pulls on the leg-foot segment with a magnitude of +88 N (as we established earlier), the leg-foot segment must pull on the capsule (legfoot-on-capsule [LfC]) with an equivalent
HR (110 N)
BR (110 N)
HR (110 N)
RH (110 N)
MsH (110 N)
RB (110 N)
BR (110 N)
Figure 1–30 The tensile forces of the pull of hand-on-rope (HR) and the pull of the cement block on the rope (BR) produce two forces of equal magnitude (110 N) that result in 110 N of tension within and throughout the rope.
Figure 1–31 Equilibrium of the man will be achieved when the force of the rope-on-hand (RH) reaches the magnitude of muscles-on- hand (MsH). Rope-on-hand will not reach the 110 N magnitude needed to establish equilib- rium until the tension in the initially slack rope reaches that magnitude as the man accel- erates away from the block.
force of –88 N (Fig. 1–33). Two questions can be raised around the assumption that there is 88 N of tension in the capsule: (1) Does the magnitude of tension reach 88 N in the capsule immediately, and (2) can the injured capsule (and ligaments) withstand 88 N of tension? Case Applica- tion 1-1 applies the concepts from the example of tension in the rope to John’s joint capsule.
MsH (200 N)
LfC (88 N) CLf (88 N)
Figure 1–33 The pull of the capsule-on-legfoot (CLf) must have a concomitant reaction force of legfoot-on-capsule (LfC) that is an 88-N tensile force on the joint capsule.
Figure 1–32 If the rope cannot with- stand the tensile forces placed on it, it will break. Once the rope breaks, the force of muscles-on-hand (MsH) is unopposed, and the man will accelerate backward.
CASE APPLICATION
Tension in the Knee
Joint Capsule case 1–1
The reaction forces of capsule-on-legfoot (CLf) and legfoot- on-capsule (LfC) have a magnitude of 88 N (see Fig. 1–33).
Side-bar: Vectors LfC and CLf should be co-linear in the figure but are separated for clarity.
Legfoot-on-capsule is a tensile vector. Tension can occur in a passive structure only if there are two pulls on the object.
Therefore, there must be a second tensile vector (of +88 N) applied to the capsule from something touching the capsule at the other end. The second tensile vector, therefore, must be femur-on-capsule (FC) (Fig. 1–34), where the tensile vectors are co-linear. The magnitudes of capsule-on-legfoot, legfoot-on-capsule, and femur-on-capsule are equivalent because capsule-on-legfoot is part of the same linear force system with weightboot-on-legfoot and gravity-on-legfoot (see Fig. 1–26C) and because vectors legfoot-on-capsule and femur-on-capsule are part of the same linear force sys- tem. The sum of the forces in both linear force systems is zero because it is assumed that no movement is occurring.
• Tensile forces are co-linear, coplanar, and applied to the same object; therefore, tensile vectors are part of the same linear force system.
• Tensile forces applied to a flexible or rigid structure of homogenous composition create the same tension at all points along the long axis of the structure in the ab- sence of friction; that is, tensile forces are transmitted along the length (long axis) of the object.
Joint Distraction
Joint capsule and ligaments are not necessarily in a constant state of tension. In fact, if John started out with his leg-foot segment on the treatment table, there would effectively be no tension in his capsule or ligaments because the sum of the forces on the leg-foot segment from the “contacts” of gravity (gravity-on-legfoot) and the treatment table (table- on-legfoot) (Fig. 1–35) would be sufficient for equilibrium (ΣF = 0). Although both the capsule and the weight boot are still attached to the leg-foot segment, the magnitudes of pull would be negligible (too small to include in the space diagram).
ConceptCornerstone 1-7
Tension and Tensile Forces
• Tensile forces (or the resultants of tensile forces) on an object are always equal in magnitude, opposite in direction, and applied parallel to the long axis of the object.
A situation similar to the leg-foot segment on the treat- ment table would exist if John’s foot were supported by someone’s hand while his leg-foot segment is in the vertical position. In Figure 1-36, the hand is pushing up (+88 N) on the leg-foot segment (hand-on-legfoot segment [HLf]) with a magnitude equivalent to the pull of gravity and the weight boot (–88 N).
Side-bar: The hand-on-legfoot vector is shown to one side of gravity-on-legfoot and weightboot-on-legfoot for clar- ity, but assume that the supporting hand is directly below the weight boot.
The magnitude of the pull of the capsule (and ligaments) on the leg-foot segment would be negligible as long as the hand-on-legfoot had a magnitude equal and opposite to that of gravity-on-legfoot and weightboot-on-legfoot. As the upward support of the hand is taken away, however, there would be a net unbalanced force down on the leg-foot seg- ment that would cause the leg-foot segment to accelerate away from the femur. The pull or movement of one bony segment away from another is known as joint distraction.10
As the upward push of the hand decreases and the leg-foot segment moves away from the femur, the capsule will be- come increasingly tensed. The magnitude of acceleration of the leg-foot segment will be directly proportional to the un- balanced force and indirectly proportional to the combined mass of the leg-foot segment and weight boot. However, the unbalanced force is difficult to quantify because it is con- stantly changing. Although the increase in capsular tension occurs concomitantly with the reduction in hand support, the two forces are not equivalent in magnitude because the leg-foot segment must move away from the femur for the capsule to get tighter; that is, there must be a net unbal- anced force on the leg-foot segment to create the movement that causes the capsule to get tighter.
The Continuing Exploration box titled “Reactions to Leg-Foot Segment Forces” presented the calculation of acceleration of the leg-foot segment at one point in time (a static rather than dynamic analysis). However, the con- cepts are more important than the calculations because the magnitudes of the weight of a limb, the support of the hand, and the tension in the capsule are generally unknown in a true clinical situation.
Figure 1–34 The tensile forces of legfoot-on-capsule (LfC) and femur-on-capsule (FC) are shown acting on the capsule.
Figure 1–36 As long as the 88-N force on the leg-foot segment from gravity (GLf) and the weight boot (WbLf) are supported by an equal upward force from the hand (HLf), the tension in the capsule and ligaments will be zero (or negligible).
Figure 1–35 The forces of table-on-legfoot (TLf) and gravity- on-legfoot (GLf) in this position are sufficient for equilibrium of the leg-foot segment, with zero (or negligible) tension in the knee joint capsule.
FC (88 N) CLf (88 N)
LfC (88 N)
Continuing Exploration 1-6:
Acceleration in Joint Distraction
The leg-foot segment and weight boot together weigh 88 N.
To calculate acceleration (a = Funbal ÷ m), however, the mass (not just the weight) of the leg-foot segment and weight boot must be known. Recalling that 1 N is the amount of force needed to accelerate 1 kg at 1 m/sec2,
GLf (48 N) TLf (48 N)
HLf (88 N) GLf (48 N)
WbLf (40 N)
In the human body, the acceleration of one or both seg- ments away from each other in joint distraction (the dynamic phase) is very brief unless the capsule and ligaments (or mus- cles crossing the joint) fail. John’s leg-foot segment will not accelerate away from the femur for very long before the distraction forces applied to the adjacent joint segments (leg-foot and femur) are balanced by the tensile forces in the capsule. Given that John is still relaxed as the weight boot hangs on his leg-foot segment (we have not asked him to do anything yet), the check to joint distraction (the pull of grav- ity and the weight boot) is the tension in the capsule (and ligaments). John presumably has a ligamentous injury that is likely to cause pain with tension in these pain-sensitive con- nective tissues. If the distraction force remains, the capsule and ligaments may fail either microscopically or macroscop- ically (see Chapter 2). For now, we can prevent this problem by putting the supporting hand back under the weight boot.
If the upward push of the hand is sufficient, the tensile forces on the ligaments can be completely eliminated.
When the hand in Figure 1–36 is no longer in contact with the leg-foot segment and the tension in the capsule reaches 88 N, the leg-foot segment will stop accelerating away from the femur and will reach equilibrium.
Distraction Forces
The resultant pull of gravity and the weight boot on the leg-foot segment (composed into a single vector) can be re- ferred to as a distraction force3or joint distraction force.
A distraction force is directed away from the joint surface to which it is applied, is perpendicular to its joint surface, and leads to the separation of the joint surfaces.
Side-bar: It is important to note that here the term distrac- tion refers to separation of rigid non-deformable bones.
Distraction across or within a deformable body is more complex and will be considered in Chapter 2.
A joint distraction force cannot exist in isolation; joint surfaces will not separate unless there is a distraction force applied to the adjacent segment in the opposite direction.
As the leg-foot segment is pulled away from the femur, any tension in the capsule created by the pull of the leg-foot segment on the capsule results in a second tensile vector in the capsule (femur-on-capsule). If the femur pulls on the capsule (see Fig. 1–34), then the capsule must concomi- tantly pull on the femur. If there is no opposing force on the femur, the net unbalanced downward force on the leg- foot segment will be transmitted through the capsule to the femur; the femur will also accelerate downward as soon as any appreciable tension is developed in the capsule. If the femur accelerates downward with the same magnitude of acceleration as the leg-foot segment, the joint surfaces will not separate any farther than was required to initiate move- ment of the femur. Although we did not set the femur in equilibrium (did not stabilize the femur) in Case Applica- tion 1-1, there must be a force applied to the femur that is opposite in direction to capsule-on-femur for there to be effective joint distraction. Joint distraction can occur only when the acceleration of one segment is less than (or in a direction opposite to) the acceleration of the adjacent seg- ment, resulting in a separation of joint surfaces.
weight (in newtons or equivalently in kg-m/sec2) is mass (in kilograms) multiplied by the acceleration of gravity, or:
W =(m)(9.8 m/sec2)
Solving for mass, a weight of 88 N is equivalent to 88 kg-m/sec2÷ 9.8 m/sec2=8.97 kg. Consequently, the leg-foot segment and weight boot together have a mass of approximately 9 kg. Assigning some arbitrary values, assume that a downward force of –88 N is offset in this static example by an upward push of the supporting hand of +50 N and capsular tensile force of +10 N. The net unbalanced force on the leg-foot segment (–88 +50 +10) is –28 N. Therefore:
a =–3.11 m/sec2
Continuing Exploration 1-7:
Stabilization of the Femur
Because our interest is primarily in John Alexander’s leg- foot segment and secondarily in the injured knee joint capsule, the source of stabilization of the femur (the other joint distraction force) was not a necessary component of our exploration. However, the principles established thus far will allow us to identify that distraction force.
In Figure 1–37, weightboot-on-legfoot and gravity-on- legfoot forces are composed into a single resultant distrac- tion force (GWbLf) of –88 N, with the leg-foot segment once again unsupported. Gravity/weightboot-on-legfoot creates an 88 N tensile force in the capsule that creates a pull of the capsule on the femur (CF) with an equal magnitude of 88 N (see Fig. 1–37). A net distractive force of +88 N ap- plied to the femur is necessary to stabilize the femur and create tension in the capsule.
The femur is contacted by both gravity-on-femur (GF) and treatment-table-on-femur (TF) (see Fig. 1–37). To determine the net force acting on the femur, we must estimate the mass or weight of that segment. John weighs approximately 734 N, and his thigh constitutes approxi- mately 10.7% of his body weight.1 His thigh, conse- quently, is estimated to weigh approximately 78 N. With the magnitudes of capsule-on-femur (–88 N) and gravity- on-femur (–78 N) known, it appears that the magnitude of table-on-femur should be the sum of the magnitudes of capsule-on-femur and gravity-on-femur (but opposite in direction). However, these vectors are not in a linear force system because they are not co-linear. Rather, they are parallel forces. Although we will tackle composition (and the effects) of parallel forces in more detail later, we can use the same shorthand system here that we used to com- pose two gravitational vectors earlier in the chapter.
Because both gravity-on-femur and capsule-on-femur (see Fig. 1–37) are vertically downward, the resultant of these two forces would be a new downward force with the
Continued a= −28 kg
9 kg-m/sec2