Chapter 4 Problems

A.

Using the percent body mass values for each segment given below the homework, we can calculate the $x$ component of COM as

$$\footnotesize \frac{0.81(8.1) + 0.58(2.8) + 0.45(2.2) + 0.6(2.8) + 0.53(2.2) + 0.57(49.7) \ + 0.41(10.0) + 0.42(4.65) + 0.39(1.45) + 0.36(10.0) + 0.22(4.65) + 0.08(1.45)}{8.1 + 2.8 + 2.2 + 2.8 + 2.2 + 49.7 + 10.0 + 4.65 + 1.45 + 10.0 + 4.65 + 1.45} = 0.52$$

and the $y$ component of COM as

$$\footnotesize \frac{1.28(8.1) + 1.19(2.8) + 1.35(2.2) +1.25(2.8) + 1.37(2.2) + 1.05(49.7) + 0.76(10.0) + 0.24(4.65) + 0.24(1.45) + 1.09(10.0) + 1.32(4.65) + 1.45(1.45)}{8.1 + 2.8 + 2.2 + 2.8 + 2.2 + 49.7 + 10.0 + 4.65 + 1.45 + 10.0 + 4.65 + 1.45} = 1.04.$$

Therefore, the COM is located at $$\begin{pmatrix}0.52 \\ 1.04\end{pmatrix}.$$

B.

In this position, she is not stable. This is because her center of mass is not above her base of support, and is instead displaced to the positive $x$ direction. In order to recover to a position of stability, she would need an external force applied in the necagive $x$ direction. Since her left foot is the only part of her body contacting an object to produce force against, the only way to generate a force in the negative $x$ direction would be to apply a force by her foot on the beam $F_\text{Foot on Beam}$ in the positive $x$ direction, creating a third law pair force to push herself back towads the negative $x$, $F_\text{Beam on Foot}$. There are several dificulties with this.

Firstly, in order for $F_\text{Foot on Beam}$ to be in the positive $x$ direction, it should come from knee flexion or hip extension because these are the movements which translate the foot posteriorly relative to COM. Since knee flexion also translates the foot superiorly relative to COM, it will decrease the normal force against the beam, reducing the ability to create our desired horizontal force (friction). Therefore, most of this movement should come from hip extension. Since Alicia’s hip is already near full extension in the photo, this strategy is unlikely to work.

Secondly, any success in producing a force $F_\text{Beam on Foot}$ in the negative $x$ direction will create a negative torque on her body, accelerating her angular velocity into the page. Since this angular velocity into the page is only a net angular velocity, she can stabilize the more important segments of her body (supporting leg and trunk) by increasing the angular velocity of segments which aren’t supporting her at the moment (swinging leg and arms). While her shoulders may allow her to continue rotating her shoulders into the page, her right hip is unlikely to flex much farther than its current position. As this continues, her increasing angular velocity will eveually affect her supporting leg and cause a fall.