KINETICS: FORCES AND MOMENTS OF FORCE - Biomechanics and Motor Control of Human Movement - page 115

1. F x = ma x ,
R x 1 = 1 . 16 × 9 . 07 = 10 . 52 N

2. F y = ma y ,

R y 1 − 1 . 16 g = m( − 6 . 62 )

R y 1 = 1 . 16 × 9 . 8 − 1 . 16 × 6 . 62 = 3 . 69 N

3. At the COM of the foot, M = I 0 α ,
M 1 − R x 1 × 0 . 0985 − R y 1 × 0 . 0195 = 0 . 0105 × 21 . 69
M 1 = 0 . 0105 × 21 . 69 + 10 . 52 × 0 . 0985 + 3 . 69 × 0 . 0195

= 0 . 23 + 1 . 04 + 0 . 07 = 1 . 34 N · m

Discussion
1. The horizontal reaction force of 10.52 N at the ankle is the cause of the
horizontal acceleration that we calculated for the foot.
2. The foot is decelerating its upward rise at the end of lift-off. Thus,
the vertical reaction force at the ankle is somewhat less than the static
gravitational force.
3. The ankle muscle moment is positive, indicating net dorsiflexor activity
(tibialis anterior), and most of this moment (1.04 out of 1.34 N · m)
is required to cause the horizontal acceleration of the foot's center of
gravity, with very little needed (0.23 N

·

m) to angularly accelerate the

low moment of inertia of the foot.

Example 5.3 (see Figure 5.8). For the same instant in time, calculate the
muscle moments and reaction forces at the knee joint. The leg segment was
43.5 cm long.

m = 0 . 0465 × 80 = 3 . 72 kg

ρ 0 = 0 . 302 × 0 . 435 = 0 . 131 m

3 . 72 ( 0 . 131 ) 2 m 2 I 0 = = 0 . 0638 kg ·

36 . 9 rad/s 2 α =

From Example 5.2, R x 1 = 10 . 52 N, R y 1 = 3 . 69 N, and M 1 = 1 . 34 N · m.

1. F x = ma x ,

R x 2 − R x 1 = ma x

R x 2 = 10 . 52 + 3 . 72 ( − 0 . 03 ) = 10 . 41 N