Lubrication of Joints - Biofluidmechanics: An Introduction to Fluid Mechanics Macrocirculation and Microcirculation - page 305

FIGURE 11.5 Diagram of a hip joint during running. Adapted from Ozkaya and Nordin (1999).

femoral head and a horizontal plane is 40 . The angle between the femur and a horizontal plane
is 75 . The hip abductor muscle attaches at the junction of the femoral head and the femur at an
angle of 75 . Assume that the weight of the leg is 15% of the total body weight and the reaction
force at the floor is 340% of the body weight (because of running). Assume that the direct length
between the point O (free-body diagram) and the femoral head is 9 cm. Assume that the weight
of the leg acts a distance of 35 cm from point O and that the reaction force acts at a distance of
88 cm from point O. The person has a weight of 800 N.

Solution
First, let us draw a free-body diagram of the leg.
In Figure 11.6 , W L is the weight of the leg, W is the reaction weight from the floor, F A is the
force of the abductor muscle, and F H is the force exerted by the hip joint. The angles are not
shown on this figure.
Writing the equations of motion in vector form, we have
P F x 5

75 Þ 2 F H cos 0 5 F A cos ð ðφÞ

X F y 5

75 Þ 2 W L 1 W 0

52 F H sin

ðφÞ 1 F A sin ð

P M O 5

40 ÞÞðF H cos 40 ÞÞðF H sin 0

5 ðð

9cm Þ sin ð ðφÞÞ 2 ðð 9cm Þ cos ð ðφÞÞ

75 ÞÞW L 1 ðð 75 ÞÞW

2 ðð

35 cm Þ cos ð 88 cm Þ cos ð

Using the third equation,
ð 5 : 785 cm ÞF H cos ðφÞ 2 ð 6 : 89 cm ÞF H sin ðφÞ 5 9 : 059 cm ð 0 : 15 800 N Þ 2 22 : 77 cm ð 3 : 4 800 N Þ

60 ; 864 Ncm

52

From the first and second equations,

F H cos ðφÞ 5 0 : 259 F A

F H sin ðφÞ 5 0 : 966 F A 2 0 : 15 800 N 3 : 4 800 N 0 : 966 F A 1 2600 N

1 5