11.67 •• BIO Downward-Facing Dog. The yoga exercise “DownwardFacing Dog” requires stretching your hands straight out above your head and bending down to lean against the floor. This exercise is performed by a 750 N person as shown in Fig. P11.67. When he bends his body at the hip to a 90° angle between his legs and trunk, his legs, trunk, head, and arms have the dimensions indicated. Furthermore, his legs and feet weigh a total of 277 N, and their center of mass is 41 cm from his hip, measured along his legs. The person’s trunk, head, and arms weigh 473 N, and their center of gravity is 65 cm from his hip, measured along the upper body. (a) Find the normal force that the floor exerts on each foot and on each hand, assuming that the person does not favor either hand or either foot. (b) Find the friction force on each foot and on each hand, cassuming that it is the same on both feet and on both hands (but not necessarily the same on the feet as on the hands). [Hint: First treat his entire body as a system; then isolate his legs (or his upper body).]

To solve this problem, we can use the principles of static equilibrium. In this position, the person is not accelerating vertically, so the sum of forces in the vertical direction is zero, and the sum of torques (moments) about any point is zero.

**(a) Normal Force:**

Let's consider the entire body as a system. The person's weight ($W_{\text{total}} = 750 \, \text{N}$) acts downward from the center of mass (C) of the entire body.

\[ W_{\text{total}} = W_{\text{legs}} + W_{\text{trunk}} \]

1. **Weight of Legs ($W_{\text{legs}}$):**

\[ W_{\text{legs}} = 277 \, \text{N} \]

The center of mass of the legs ($C_{\text{legs}}$) is 41 cm from the hip.

2. **Weight of Trunk ($W_{\text{trunk}}$):**

\[ W_{\text{trunk}} = 473 \, \text{N} \]

The center of gravity of the trunk ($C_{\text{trunk}}$) is 65 cm from the hip.

Now, we can find the normal force ($N$) on each foot and each hand.

\[ N = \frac{W_{\text{legs}}}{2} + \frac{W_{\text{trunk}}}{2} \]

Substitute the given values and solve for $N$.

**(b) Friction Force:**

In the downward-facing dog position, the friction force ($f$) prevents slipping. We'll consider the friction on the feet and the hands separately.

1. **Friction on Feet ($f_{\text{feet}}$):**

\[ f_{\text{feet}} = \mu_{\text{feet}} \cdot N_{\text{feet}} \]

2. **Friction on Hands ($f_{\text{hands}}$):**

\[ f_{\text{hands}} = \mu_{\text{hands}} \cdot N_{\text{hands}} \]

The coefficient of friction ($\mu$) is not provided, so it's a symbolic variable in the equations. You would need the specific value to find the actual friction force.

This approach considers the entire body as a system and then breaks down the forces for each part.