Section 1.8
Section 1.8 The Back
From Eq. 1.18, we obtain FR cos θ 1.59W cos 71◦ 0.52W therefore, 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 whose mass is 70 kg and weight is 9.8 × 70 686 N (154 lb). The force on the hip joint is 1625 N (366 lb).
1.7.1 Limping
Persons who have an injured hip limp by leaning toward the injured side as they step on that foot (Fig. 1.15). As a result, the center of gravity of the body shifts into a position more directly above the hip joint, decreasing the force on the injured area. Calculations for the case in Fig. 1.15 show that the muscle force Fm 0.47W and that the force on the hip joint is 1.28W
(see Exercise 1-11). This is a significant reduction from the forces applied during a normal one-legged stance.
1.8The Back When the trunk is bent forward, the spine pivots mainly on the fifth lumbar vertebra (Fig. 1.16a). We will analyze the forces involved when the trunk is bent at 60◦ from the vertical with the arms hanging freely. The lever model representing the situation is given in Fig. 1.16.
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 weight of the head and arms is represented by W2 suspended at the end of the lever arm. The erector spinalis muscle, shown as the connection D-C attached at a point two-thirds up the spine, maintains the position of the back. The angle between the spine and this muscle is about 12◦. For a 70-kg man, W1 and W2 are 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
Chapter 1 Static ForcesFIGURE 1.15 Walking on an injured hip.
the compressional force of the fifth lumbar vertebra is 2230 N (500 lb). If, in addition, the person holds a 20-kg weight in his hand, the force on the muscle is 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 this point. It is evident too that the position shown in the figure is not the recommended way of lifting a weight.
From Eq. 1.18, we obtain FR cos θ 1.59W cos 71◦ 0.52W therefore, 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 whose mass is 70 kg and weight is 9.8 × 70 686 N (154 lb). The force on the hip joint is 1625 N (366 lb).
1.7.1 Limping
Persons who have an injured hip limp by leaning toward the injured side as they step on that foot (Fig. 1.15). As a result, the center of gravity of the body shifts into a position more directly above the hip joint, decreasing the force on the injured area. Calculations for the case in Fig. 1.15 show that the muscle force Fm 0.47W and that the force on the hip joint is 1.28W
(see Exercise 1-11). This is a significant reduction from the forces applied during a normal one-legged stance.
1.8The Back When the trunk is bent forward, the spine pivots mainly on the fifth lumbar vertebra (Fig. 1.16a). We will analyze the forces involved when the trunk is bent at 60◦ from the vertical with the arms hanging freely. The lever model representing the situation is given in Fig. 1.16.
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 weight of the head and arms is represented by W2 suspended at the end of the lever arm. The erector spinalis muscle, shown as the connection D-C attached at a point two-thirds up the spine, maintains the position of the back. The angle between the spine and this muscle is about 12◦. For a 70-kg man, W1 and W2 are 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
Chapter 1 Static ForcesFIGURE 1.15 Walking on an injured hip.
the compressional force of the fifth lumbar vertebra is 2230 N (500 lb). If, in addition, the person holds a 20-kg weight in his hand, the force on the muscle is 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 this point. It is evident too that the position shown in the figure is not the recommended way of lifting a weight.