Section 1.6 The Elbow
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FIGURE 1.10 Motion of the lever arms in a Class 1 lever.
A force slightly greater than what is required to balance the load will lift it. As the point at which the force is applied moves through a distance L2, theload moves a distance L1 (see Fig. 1.10). The relationship between L1 andL2, (see Exercise 1-2) is given by L1 d1.
(1.8)
L2 d2 The ratio of velocities of these two points on a moving lever is likewise given by v1 d1.
(1.9)
v2
d2 Here v2 is the velocity of the point where the force is applied, and v1 is thevelocity of the load. These relationships apply to all three classes of levers.
Thus, it is evident that the excursion and velocity of the load are inverselyproportional to the mechanical advantage.
1.6 The Elbow
In our analysis of the elbow, we will consider the action of only these twomuscles. This is a simplification, as many other muscles also play a role inelbow movement. Some of them stabilize the joints at the shoulder as theelbow moves, and others stabilize the elbow itself.
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Chapter 1 Static Forces FIGURE 1.11 The elbow.
FIGURE 1.12 (a) Weight held in hand. (b) A simplified drawing of (a).
Section 1.6 The Elbow
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FIGURE 1.13 Lever representation of Fig. 1.12.
Figure 1.12a shows a weight W held in the hand with the elbow bent at a 100◦ angle. A simplified diagram of this arm position is shown in Fig. 1.12b.
The dimensions shown in Fig. 1.12 are reasonable for a human arm, but theywill, of course, vary from person to person. The weight pulls the arm downward. Therefore, the muscle force acting on the lower arm must be in the updirection. Accordingly, the prime active muscle is the biceps. The position ofthe upper arm is fixed at the shoulder by the action of the shoulder muscles.
We will calculate, under the conditions of equilibrium, the pulling force exerted by the biceps muscle and the direction and magnitude of the reactionforce Fr at the fulcrum (the joint). The calculations will be performed by considering the arm position as a Class 3 lever, as shown in Fig. 1.13. The x- andy-axes are as shown in Fig. 1.13. The direction of the reaction force Fr shownis a guess. The exact answer will be provided by the calculations.
In this problem we have three unknown quantities: the muscle force Fm, the reaction force at the fulcrum Fr, and the angle, or direction, of this forceφ. The angle θ of the muscle force can be calculated from trigonometric considerations, without recourse to the conditions of equilibrium. As is shown inExercise 1-3, the angle θ is 72.6◦.
For equilibrium, the sum of the x and y components of the forces must each be zero. From these conditions we obtain x components of the forces :
Fm cos θ Fr cos φ
(1.10)
y components of the forces :
Fm sin θ W + Fr sin φ
(1.11)
These two equations alone are not sufficient to determine the three unknownquantities. The additional necessary equation is obtained from the torque conditions for equilibrium. In equilibrium, the torque about any point in Fig. 1.13
must be zero. For convenience, we will choose the fulcrum as the point forour torque balance.
Chapter 1 Static Forces y component of the muscle force. Since the reaction forceFr acts at the fulcrum, it does not produce a torque about this point.
Using the dimensions shown in Fig. 1.12, we obtain 4 cm × Fm sin θ 40 cm × W Fm sin θ 10W
(1.12)
Therefore, with θ 72.6◦, the muscle force Fm is Fm 10 W 10.5W
(1.13)
0.954 With a 14-kg (31-lb) weight in hand, the force exerted by the muscle is Fm 10.5 × 14 × 9.8 1440 N (325 lb) If we assume that the diameter of the biceps is 8 cm and that the muscle can produce a 7 × 106 dyn force for each square centimeter of area, the armis capable of supporting a maximum of 334 N (75 lb) in the position shown inFig. 1.13 (see Exercise 1-4).
The solutions of Eqs. 1.10 and 1.11 provide the magnitude and direction of the reaction force Fr. Assuming as before that the weight supported is14 kg, these equations become 1440 × cos 72.6 Fr cos 1440 × sin 72.6 14 × 9.8 + Fr sin φ
(1.14)
or Fr cos φ 430 N Fr sin φ 1240 N
(1.15)
Squaring both equations, using cos2 φ + sin2 φ 1 and adding them, weobtain F 2 r 1.74 × 106 N2 or Fr 1320 N (298 lb)
(1.16)
From Eqs. 1.14 and 1.15, the cotangent of the angle is cot φ 430 0.347
(1.17)
1240
and
φ 70.9◦
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Exercises 1-5, 1-6, and 1-7 present other similar aspects of biceps mechanics.
In these calculations we have omitted the weight of the arm itself, but thiseffect is considered in Exercise 1-8. The forces produced by the triceps muscleare examined in Exercise 1-9.
Our calculations show that the forces exerted on the joint and by the muscle are large. In fact, the force exerted by the muscle is much greater than theweight it holds up. This is the case with all the skeletal muscles in the body.
They all apply forces by means of levers that have a mechanical advantageless than one. As mentioned earlier, this arrangement provides for greaterspeed of the limbs. A small change in the length of the muscle produces arelatively larger displacement of the limb extremities (see Exercise 1-10). Itseems that nature prefers speed to strength. In fact, the speeds attainable atlimb extremities are remarkable. A skilled pitcher can hurl a baseball at aspeed in excess of 100 mph. Of course, this is also the speed of his hand at thepoint where he releases the ball.
1.7 The Hip Fm. When a person stands erect, the angle of this force is about 71◦with respect to the horizon. WL represents the combined weight of the leg,foot, and thigh. Typically, this weight is a fraction (0.185) of the total bodyweight W (i.e., WL 0.185 W ). The weight WL is assumed to act verticallydownward at the midpoint of the limb.
We will now calculate the magnitude of the muscle force Fm and the force FR at the hip joint when the person is standing erect on one foot as in a slowwalk, as shown in Fig. 1.14. The force W acting on the bottom of the lever isthe reaction force of the ground on the foot of the person. This is the forcethat supports the weight of the body.
From equilibrium conditions, using the procedure outlined in Section 1.6, we obtain Fm cos 71◦ − FR cos θ 0 (x components of the force 0)
(1.18)
Fm sin 71◦ + W − WL − FR sin θ 0 (y components of the force 0)
(1.19)
(FR sin θ) × 7 cm + WL × 10 cm −W × 18 cm 0 (torque about point A 0)
(1.20)
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Chapter 1 Static Forces FIGURE 1.14 (a) The hip. (b) Its lever representation.
Since WL 0.185 W, from Eq. 1.20 we have FR sin θ 2.31W Fm 1.50W .59W
(1.21)
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From Eq. 1.18, we obtain FR cos θ 1.59W cos 71◦ 0.52W θ tan−1 4.44 77.3◦ and FR 2.37W
(1.22)
This calculation shows that the force on the hip joint is nearly two and onehalf times the weight of the person. Consider, for example, a person whosemass is 70 kg and weight is 9.8 × 70 686 N (154 lb). The force on the hipjoint is 1625 N (366 lb).
1.7.1 Limping Fm 0.47W and that the force on the hip joint is 1.28(see Exercise 1-11). This is a significant reduction from the forces appliedduring a normal one-legged stance.
1.8 The Back
The pivot point A is the fifth lumbar vertebra. The lever arm AB represents the back. The weight of the trunk W1 is uniformly distributed along the back;its effect can be represented by a weight suspended in the middle. The weightof the head and arms is represented by W2 suspended at the end of the leverarm. The erector spinalis muscle, shown as the connection D-C attached at apoint two-thirds up the spine, maintains the position of the back. The anglebetween the spine and this muscle is about 12◦. For a 70-kg man, W1 and W2are typically 320 N (72 lb) and 160 N (36 lb), respectively.
Solution of the problem is left as an exercise. It shows that just to hold up the body weight, the muscle must exert a force of 2000 N (450 lb) and
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Chapter 1 Static Forces FIGURE 1.15 Walking on an injured hip.
the compressional force of the fifth lumbar vertebra is 2230 N (500 lb). If, inaddition, the person holds a 20-kg weight in his hand, the force on the muscleis 3220 N (725 lb), and the compression of the vertebra is 3490 N (785 lb)(see Exercise 1-12).
This example indicates that large forces are exerted on the fifth lumbar vertebra. It is not surprising that backaches originate most frequently at thispoint. It is evident too that the position shown in the figure is not the recommended way of lifting a weight.
