## Examples from Weight Lifting

Weight lifting involves the rotation of a body segment against resistance. Because the exercise is done slowly (about one repetition per second), the inertial effects are neglected. We next present examples from the static analysis of weight lifting.

Example 6.4. Deltoids. Deltoids are a shoulder muscle group that is located on the upper side of the arms. Deltoids originate at the bones of the shoulder (clavicle and scapula) and end at the outer midsection of Figure 6.5a,b. A man performing standing lateral raise (a). The free-body diagram of his arm is shown in (b).

humerus. The primary function of this muscle group is to abduct the arms. Standing lateral raises (Fig. 6.5) is a major exercise for this muscle group. In performing this exercise, select a weight that allows you to warm up and learn the proper movement. Stand with chest out, back straight, and chin level. Starting with hands at your side raise the dumbbells upward to shoulder height, with elbows slightly bent. Lower the weight slowly to the starting position. Repeat movement.

Static Analysis of the Exercise: The free-body diagram of the arm is shown in Fig. 6.5b. The condition of zero moment at the center of rotation of the shoulder joint (point A) requires that

Fd • d - M\ g (L/2) sin 9 - m g L sin 9 = 0 (6.8a)

in which Fd denotes the force produced by the deltoids, d is the lever arm of this force with respect to point A, m\ is the weight of the arm (including the forearm and hand), L is the length of the arm, m is the mass of the weight carried, and 6 is the angle of inclination of the arm with the vertical axis. Rearranging this equation we obtain:

According to the data presented in Appendix 2, one set of upper arm, forearm, and hand constitute on the average 6.5% of the body weight of a young adult male. For a man who weighs 79 kg, this would amount to 5 kg; thus, mi = 5 kg. If the man is lifting 20-lb dumbbells, then m = 9 kg. If his upper limb is 73 cm long, the moment created by the deltoid muscle group is

In general, the moment arm of the deltoids will change with the angle 6, but for simplicity let us assume that d = 7.5 cm for all angles under consideration. Then,

According to this equation, the maximum force carried by the deltoids occur when the arms are aligned horizontally (6 = ^/2). It is interesting to note that a muscle that weighs 1 to 2 kg can produce a force in excess of 200 kg in a young male. Also note that the moment created by the weights at the shoulder increases with increasing arm length. Thus, at first glance, it may appear as if the longer the arm, the higher must be the force produced by the deltoids. However, the moment arm d is probably larger in people with longer arms, and thus may counterbalance any increase in the muscle force caused by longer arm length.

Let us next consider the total joint force acting at the shoulder joint. Assuming that the deltoid muscle force acts in the direction of the arm, we have the following relations:

in which R1 and R2 represent, respectively, the horizontal and vertical components of the total joint force Fj. Rearranging Eqn. 6.9 and substituting the input parameter values, one obtains

Thus, the total shoulder joint force is largely a compressive force that is comparable in magnitude to the force produced by the deltoids. It is likely that much of this force is resisted by the bone of the upper arm humerus. Passive soft tissue structures such as ligaments have low resistance to compression.

Example 6.5. Erector Spinae. Determine the tensile force produced by the erector spinae for the body positions shown in Fig. 6.6. Assume that the exercise is performed slowly so that equations of static equilibrium hold. The length L between the tip of the head of the athlete and his hip joint is 82 cm. The length Lc between the center of mass of the upper body and the hip joint is 44 cm. The upper body weighs 37 kg. The lever arm d of the erector spinae is 4 cm and is largely insensitive to the angle between the upper and lower body.

Solution: Consider the free-body diagram of the upper body shown in Fig. 6.6b. The resultant moment acting on the upper body must be equal to zero. If we take moments with respect to the point of application of the total joint force in the hip joint, following equation is obtained:

Fe • d - m g Lc sin 9 = 0 => Fe = m g (Lc/d) sin 9

in which Fe is the force produced by the erector spinae. Substituting the input parameter values into this equation, we find

These results suggest that erector spinae is capable of generating forces that are four to five times the body weight. Figure 6.6a,b. A man performing back curls to strengthen his back muscle erector spinae (a). The free-body diagram of his upper body is shown in (b).

The static analysis presented here can be used to evaluate the forces involved in a large number of weight-lifting exercises. Some of the body dimensions required in the analysis can be obtained by direct measurement. Other input parameters such as the weight of a forearm or upper body can be evaluated by knowing the body weight and using it in the tables given in Appendix 2. Admittedly, the input data thus obtained will be approximations to direct measurements. However, the procedure gives a starting point for "back-of-the-envelope" computations. The lever arms of various muscle groups that move a body segment may be the hardest to measure. The data obtained by cadaver studies and by X-ray provide estimates for some of the important lever arms involved in movement and motion (Appendix 2). 