12.1 Mechanical Equilibrium

Static model calculations are based on a state of static mechanical equilibrium. The conditions of static mechanical equilibrium read: “a body is in static equilibrium if it experiences no accelerated linear motion (translation) and no accelerated rotation.” In static equilibrium the sum of all external forces and the sum of all external moments is equal to zero.

The symbol Σ in these equations is the mathematical sum sign. The vector sums are to be extended over all forces F ₁, F ₂, F ₃, … and all moments M ₁, M ₂, M ₃, … acting on the body. If in the two-dimensional case a moment is defined as “the product of the magnitude of the distance from the center of rotation times the magnitude of the force,” the equilibrium condition for moments can be formulated as

The designations “±” in this formula indicate that the signs of the moments have to be chosen in accordance with the sign rule for moments, that is, in dependence on the direction of rotation (clockwise or counterclockwise).

The validity of these equations can be seen immediately. If the sum of all forces were not equal to zero, a resultant force (unequal to zero) would act on the body. The body would then be linearly accelerated and, in contrast to the assumption, not be in static equilibrium. The analogous argument holds for the sum of the moments. If the sum of all moments were not equal to zero, a resultant moment (unequal to zero) would act on the body. The body would then undergo an accelerated rotation and would thus not be in static equilibrium. It follows that if a body is seen to perform no accelerated translation or rotation (in the simplest case: if the body is at rest), the sum of all forces and the sum of all moments must be zero. Additionally, in the dynamic case, when accelerated translations and/or rotations occur, the conditions of mechanical equilibrium remain valid if the sum of the forces (Eq. 12.1) is extended over the inertial force m · (−a) and the sum of the moments (Eq. 12.2) over the moment of inertia I · (−α).

Fig. 12.1 illustrates a state of mechanical equilibrium. The force from the hand onto the tray is opposite and equal to the weight of the drink and the burger (the tray is neglected); that is, the sum of the forces is equal to zero. The support provided by the hand is placed in such a way that the moment of the weight of the drink (in the case shown here, the drink is heavier than the burger) is opposite and equal to the moment of the weight of the burger. The sum of the moments equals zero as well.

In a state of mechanical equilibrium, two unknowns can be determined by using Eq. 12.1. Fig. 12.2 illustrates the calculation of an unknown moment with the example of a beam allowed to rotate about a center of rotation. Vectors F ₁ and F ₂ represent the forces exerted on the beam; F₁ and F₂ are their magnitudes. The xy-coordinate system is oriented so that both forces point in the y-direction. L₁ and L₂ are the perpendicular distances of the points of force application from the center of rotation, the “moment arms.” Initially F₁ is known; F₂ is unknown. In static equilibrium (that is, if no accelerated rotation is observed), the sum of all moments must be equal to zero. It then holds (keeping the sign convention for moments in mind) that

Thus the unknown moment L₂ · F₂ is obtained.

Fig. 12.3 illustrates the calculation of an unknown force in a condition of static mechanical equilibrium. Three forces act on a body. F ₁ and F ₂ are known; the direction and magnitude of F ₃ are unknown. In equilibrium the vector sum of all forces must be equal to zero:

To fulfill this equation, F ₃ must be oppositely directed and equal in magnitude to the sum of F ₁ and F ₂. In the graphical representation, vectors F ₁, F ₂, and F ₃ form a closed triangle. Thus the magnitude and direction of F ₃ are unequivocally determined.

12.2 Example: Calculation of the Joint Load of a Beam Balance in Static Equilibrium

To illustrate the principle of calculating joint loads in the human body by using the conditions of equilibrium, the joint load of a simple mechanical instrument, a beam balance, can be calculated ( Fig. 12.4 ). For reasons of clarity, the pointer of the beam balance (though essential to the mechanical functioning of the balance) is not shown in the illustration. In the state of static equilibrium the unknown force F, which equilibrates a gravitational force of 10 N, and the joint load G, which supports the beam at the center of rotation, are to be calculated ( Fig. 12.4a ). In a first step ( Fig. 12.4b ) the magnitude of the force F is determined using the condition of equilibrium of moments:

In a second step ( Fig. 12.4c ) G is calculated from the equilibrium of forces:

The joint load G points in the positive y-direction.

If force F, which maintains the equilibrium at the beam, is not directed in the y-direction ( Fig. 12.5 ), F is decomposed into forces F _x and F _y, and the x- and y-components of force G are determined separately. In a second step, the magnitude and direction of G are determined from its components G _x and G _y. In the setup shown in Fig. 12.5 it is assumed that the angle between the line of action of F and the beam amounts to 45°. In equilibrium of moments, F_y and G_y equal those calculated already for the setup shown in Fig. 12.4 .

The x-component of force F effects a moment equal to zero because its line of action passes through the center of rotation of the beam.

Given the angle of force F to the x-axis and with knowledge of F _y, vector F _x is determined. To do this, the missing side of the triangle formed by F, F _x, anF _y has to be constructed. If the angle (as assumed here) equals 45°, it holds that

The magnitude of the force F is determined from

In the state of static equilibrium the sum of the x-components of all forces must be equal to zero. The weight on the right side does not contribute to the sum, because the gravitational force points in the y-direction. With F_x = −20 N it follows that

The vector of the joint load G is determined from its components G _x and G _y. Its magnitude is calculated from

The angle β between the vector of the joint load and the x-axis is determined from

In the case discussed here the angle β is equal to 56.3°.

12.3 Calculation of a Joint Load in the Static Case (Example: The Elbow Joint)

The forearm is held at an angle of 90° with respect to the upper arm ( Fig. 12.6 ). The mass of 10 kg held in the hand exerts a gravitational force of 98.1 N on the hand; this value is rounded up to 100 N in the calculation below. For the time being the weight of the forearm is neglected. To maintain equilibrium the biceps muscle is activated. In relation to the xy-coordinate system shown, both external load and muscle force are parallel to the y-direction; the gravitational force W and the muscle force B have only y-components. The moment arms of the external force and the muscle force are assumed to be 20 cm and 2 cm, respectively (approximately correct, rounded values).

When writing down the equilibrium condition for the moments, attention must be paid to using the correct signs. Then

This equation yields for the required force of the biceps muscle

In the state of equilibrium, the (vector) sum of forces must equal zero:

Here the y-component of the weight W has a negative sign because the gravitational force points in the negative y-direction. The force of the biceps points in the positive y-direction and has a positive sign.

The joint load G has a magnitude of 900 N. The negative sign shows that force G points in the negative y-direction. In other words, the humerus is pressed against the ulna with a force of 900 N. It can be seen that the largest contribution to the joint load (1000 N) comes from the muscle force of the biceps.

In the example shown in Fig. 12.6 equilibrium is established solely by activation of the biceps muscle; antagonistic muscle activation is not taken into account. This is certainly a simplification because experience shows that agonistic and antagonistic muscles are usually activated simultaneously to maintain postures or perform activities. However, consideration of the force T of the triceps in addition to the force B of the biceps ( Fig. 12.7 ) immediately gives rise to the problem that the two equilibrium conditions are insufficient for unambiguous calculation of two unknown muscle forces (biceps and triceps) and the unknown joint load. To obtain an unequivocal solution an additional assumption has to be made. To explore the extent to which the joint load changes if an antagonist is activated, we assume (arbitrarily) that the force developed by the triceps is small and amounts to 25% of the force of the biceps.

To keep the calculation simple, it is further assumed that the force of the triceps is, like the force of the biceps, directed in the positive y-direction. With a moment arm of the triceps of 2 cm (approximate value) and with reference to the sign convention for moments we obtain in equilibrium of moments

From these two equations we obtain B and T:

In equilibrium of forces

we obtain for their y-components (digits after the decimal point omitted)

The joint load now amounts to 1566 N. Again, the negative sign indicates that the humerus is pressed against the ulna. The participation of the antagonist increases the joint load by a considerable amount. The force developed by the antagonist is assumed to be 25% of the force of the agonist; the joint load, however, has increased by 74% (from 900 N to 1566 N). Antagonistic muscle activation may be helpful in stabilizing postures by making them insensitive to small perturbations. If, however, a muscle is forced to carry out a movement against strong resistance from the antagonist, very high joint loading may be expected to result. This could, for example, be seen in spastic patients with overactivation of certain muscles, or in persons with Parkinson disease.

In general, the procedure for determining joint loads in a state of static equilibrium can be summarized as follows ( Fig. 12.8 ): magnitudes, directions, and points of application of all forces are marked, in so far as they are known from the outset, on the “free” body (that is, the body imagined as separated from its environment). This diagram is termed the free body diagram. In the case discussed above, the initial knowledge comprises the magnitude, direction, and point of application of the external load F ₁ and (to complete the description) the gravitational force of forearm and hand F ₂. These forces are applied at the centers of mass of the external load and of the forearm, respectively. Regarding the muscle force F ₃, only its point of application and its direction are known from anatomical observations. As far as the joint load F ₄ is concerned, only its point of application is known: due to the fact that there is only negligible friction between the articulating surfaces, this force is directed through the center of rotation. (The supporting argument for this is that a frictionless joint cannot transmit any moment. If the joint load were not directed through the center of rotation, a moment, and hence an accelerated rotational movement, would result. Static equilibrium would not exist, in contradiction of the assumption.)

In the first step, the moment of the muscle force and (with a known moment arm of the muscle) the magnitude of the muscle force F ₃ are determined from the equilibrium condition of moments. In the second step, the magnitude and direction of the joint load F ₄ are determined from the equilibrium condition of forces. The prerequisites for performing this model calculation are: (1) assuming only one muscle to be active, specifically excluding multiple or antagonistic muscle activation, (2) assuming a punctiform application of muscle and tendon forces at the bones (instead of an anatomically correct area of force application), and (3) ignoring the elastic tension of other muscles, joint capsules, or ligaments spanning the joint.

With respect to muscle forces and joint loads, the results obtained at the elbow joint are typical of many joints of the locomotor system:

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  The muscle force needed to guarantee equilibrium is, in general, much larger than the external forces acting on the body. This is due to the ratio of the moment arm of the external force compared to the moment arm of the muscles. For the long bones of the extremities it holds as a rule that moment arms of external forces are comparable to the length of the bones while moment arms of the muscles are comparable to the diameter of the bones.

  
  
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  The magnitude of the joint load is determined essentially by the muscle force; the external force makes only a small additional contribution.

  
  
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  The absolute minimum of the joint load occurs when only the agonist muscle is activated. Additional activation of antagonistic muscles increases the joint load disproportionately.

The underlying reason for the high muscle forces and joint loads is in the architecture and anatomical function of the muscles ( Fig. 12.9 ). Muscles can produce large tensile forces while the change of muscle length is limited to relatively small values. If the range of motion of a joint is large, the muscle must, for geometrical reasons, be positioned close to the joint. Its length change dL will then be small. In many cases the moment arms of external forces (or of the body weight) are much larger than the moment arms of the muscles. The ratio of the moment arms of muscles to those of the external forces requires a large muscle force to guarantee equilibrium. The large muscle forces cause large joint loads.