Question 11.3: During normal motion (walking, jogging, or running), for som...
During normal motion (walking, jogging, or running), for some instant in time, all of our body weight is supported by one leg (see Figure 11.5). Typically, the leg is not completely vertical because of the way in which the femur connects to other bones (refer to Figure 11.6, the free-body diagram). Using the following values, calculate the reaction forces that act in the hip joint during running at the instant when the body is supported by one leg. The angle between the 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.
First, let us draw a free-body diagram of the leg.
In Figure 11.6, WL is the weight of the leg, W is the reaction weight from the floor, FA is the force of the abductor muscle, and FH 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
\sum{F_{x}}= 0 = F_{A}\cos(75^{\circ}) – F_{H}\cos(\phi)
\sum{F_{y}}= 0 = – F_{H}\sin(\phi) + F_{A}\sin(75^{\circ}) – W_{L} + W
\sum{M_{O}}= 0 = ((9 cm)\sin(40^{\circ}))(F_{H}\cos(\phi)) – ((9 cm)\cos(40^{\circ}))(F_{H}\sin(\phi)) – ((35 cm)\cos(75^{\circ}))W_{L} + ((88 cm)\cos(75^{\circ}))W
Using the third equation,
(5.785 cm)F_{H}\cos(\phi) – (6.89 cm)F_{H}\sin(\phi)= 9.059 cm(0.15 \ast 800 N) – 22.77 cm(3.4 \ast 800 N) = -60,864 Ncm
From the first and second equations,
F_{H}\cos(\phi)= 0.259F_{A}
F_{H}\sin(\phi)= 0.966F_{A} – 0.15 \ast 800 N + 3.4 \ast 800 N = 0.966F_{A} + 2600 N
Substituting these two relationships into the third equation,
(5.785 cm)F_{H}\cos(\phi) – (6.89 cm)F_{H}\sin(\phi)= – 60,864 Ncm
5.785 cm(0.259F_{A}) – 6.89 cm(0.966F_{A} + 2600 N)= – 60,864 Ncm
(1.5 cm)F_{A} – (6.566 cm)F_{A} – 17,914 Ncm = – 60,864 Ncm
(-5.066 cm)F_{A} = – 42,949 Ncm
F_{A} = 8478 N
F_{H}\cos(\phi)= 0.259(8478 N)= 2196 N
F_{H}\sin(\phi)= 0.966(8478 N) + 2600 N= 10,790 N
F_{H}= \sqrt{(F_{H}\cos(\phi))^{2} + (F_{H}\sin(\phi))^{2}}= \sqrt{(2196 N)^{2} + (10,790 N)^{2}}= 11011 N
\phi= \tan^{-1} \left(\frac{10,790 N}{2196 N}\right)= 78.5^{\circ}