Biarticular Muscles
Biarticular muscles act on two joints. These muscles include some of the major muscles of the upper and lower limbs. Hamstrings, a group of three muscles, constitute an important example for biarticular muscles. These muscles originate in the ischial tuberosity of the hipbone and insert into the bones of the lower leg. When the thigh and the hip are fixed, ham strings flex the knee. They also extend the hip through the movement of the thigh. The third function of the hamstrings is to raise the trunk from a flexed position and keep it erect. Quadriceps is also a biarticular muscle group. They function mainly as the primary leg extender. The following example illustrates the effective functions of hamstrings, calfs, and quads during squats.
Example 6.8. An athlete is performing squats to strengthen knee muscles. The movement is slow enough to assume static equilibrium. Consider the fourlink system shown in Fig. 6.8 to represent the athlete during squatting. The beam representing the hip is connected to the rod representing the upper leg at the hip joint. A tensioncarrying cord representing the calf muscle connects the foot to the thigh. The quad muscle connects the thigh and the leg through a frictionless pulley mechanism representing the patella joint. The joints at H, K, and A are hinge joints. Determine the tension in hamstring, calf, and quads as a function of the angle the leg makes with the horizontal plane (6).
Solution: For simplicity we assume symmetry with respect to the horizontal plane passing through the knee joint. Let P denote the force transmitted at the hip joint to each leg. The force P is then equal to half the weight of the upper body plus the weight used for the squat. For symmetry, we assume that the legs and the feet are weightless and that the entire weight is lumped into a single weight P acting at a distance c from the hip joint H. Consider the freebody diagram of the foot shown in Fig. 6.8b. The moment created by force P about A should be equal to the moment produced by the calf muscle at A:
b sin 0 Fc + c P = 0 => Fc = c P/(b sin 0) (6.17)
in which Fc denotes the force produced by the calf muscle group. Because of symmetry in the idealized structure, the calf muscle will produce the same tension as the hamstrings:
Next, let us consider the freebody diagram of the feet and the lower legs (Fig. 6.8c). The moment of all external forces with respect to the knee joint must be equal to zero:
(L cos 6  c) P  2 dk Fh sin 0 + dq Fq = 0 (6.19)
where L represents the length of upper and lower leg. The first term on the lefthand side of Eqn. 6.19 is the moment of the P with respect to K. The second term is the moment created by the hamstring and calf muscles at the knee. Because of symmetry, the resultant of these two muscle
Figure 6.8ae. The schematic diagram of a person performing squats (a). The figure also shows the freebody diagrams (be) of body segments involved in this movement.
forces acts vertically upward. The term 2 Fh sin 0 is the resultant of the tension in hamstring and calves about the knee joint, and dk is its moment arm. The last term on the lefthand side is the counterclockwise moment created by the quad muscle group. The term dq is the moment arm of the quad tension (Fq) with respect to the knee joint.
using geometric relations, the moment arm dk can be shown to be given by the relation:
The angle 0 is related to the angle 9 and the length Lh of the hamstrings as follows:
in which h denotes the distance from the insertion of calf muscle on the femur to the center of rotation of the knee. Eliminating Lh from these equations, we obtain the following relationship for 0:
Now, that we have expressed 0 as a function of 9, we can compute the moment arms of hamstrings about point H and calves about point A.
The next step is the determination of the moment arm dq of the quad muscle group. The geometry associated with the quads is shown in Fig. 6.8d. We assume that the quads originate and insert at a distance of L/3 from the center of rotation of the knee and that the patella keeps the quads a distance u away from the femorotibial joint. We use the law of cosines successively to determine the moment arm dq:
The compressive force acting on the quads and the knee joint can be found by considering the static equilibrium of point K (Fig. 6.8e):
in which Fp represents the patellofemoral compressive force.
Let us illustrate this solution with a numerical example. Let us assume that L = 40 cm, b = 10 cm, h = 8 cm, u = 5 cm, and c = 14 cm. For body configuration at 9 = 60°:
0 = 58°, a = 45°, Fc = Fh = 1.5 P , Fq = 7.9 P, and Fp = 11.2 P
When the body goes further down so that 9 = 30°:
0 = 32°, a = 22°, Fc = Fh = 2.64 P , Fq = 22.0 P, and Fp = 40.8 P
These numerical results indicate that quads need to produce much more tension than the other two muscles involved in squatting: hamstrings and calves. Furthermore, the compressive force between the patella and the femur increases tremendously with increasing quads tension and increasing flexion of the knee. Repeated compressive force disrupts and destroys the cartilage coating of the articulating bone surfaces, leading to frictional resistance to the sliding motion of the kneecap during knee flexion and extension. This overuse injury is commonly observed among runners. The extensor mechanism of the knee is the most common site of chronic running injuries. Runningassociated pain may be caused by excessive compressive stress acting on the articular cartilage of the patellofemoral joint. Recently, research has focused on how to restore damaged cartilage to the joints of the upper and lower limbs. The recent developments in biotechnology allow the growing of cartilage cells in culture dishes in a laboratory setting. Whether injection of fibrous elements or cartilage cells grown in vitro into the damaged cartilage of a joint will result in the restoration of this important tissue remains to be seen.
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wolfgang11 months ago
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