Questions: Problem 6: (17% of Assignment Value) Even when the head is held erect, as shown in the figure, its center of mass is not directly over the principal point of support (the atlanto-occipital joint). The muscles at the back of the neck should therefore exert a force to keep the head erect. That is why your head falls forward when you fall asleep in class. Part (a) If the head has a weight of 49.9 N, calculate the force in units of newtons exerted by these muscles using the information in the figure. Part (b) What is the force in newtons exerted by the pivot on the head?

Solution Steps

• Weight of the head, \( w = 49.9 \, \text{N} \)
• Distance from the pivot to the center of gravity, \( d_{\text{cg}} = 5.0 \, \text{cm} = 0.05 \, \text{m} \)
• Distance from the pivot to the point where the muscle force acts, \( d_{\text{m}} = 2.5 \, \text{cm} = 0.025 \, \text{m} \)

To keep the head erect, the torques around the pivot point must balance. The torque due to the weight of the head (\( w \)) and the torque due to the muscle force (\( F_{\text{M}} \)) must be equal and opposite.

\[ \tau_{\text{weight}} = \tau_{\text{muscle}} \]

\[ w \cdot d_{\text{cg}} = F_{\text{M}} \cdot d_{\text{m}} \]

Rearrange the equation to solve for \( F_{\text{M}} \):

\[ F_{\text{M}} = \frac{w \cdot d_{\text{cg}}}{d_{\text{m}}} \]

Substitute the given values:

\[ F_{\text{M}} = \frac{49.9 \, \text{N} \cdot 0.05 \, \text{m}}{0.025 \, \text{m}} \]

\[ F_{\text{M}} = \frac{2.495 \, \text{N} \cdot \text{m}}{0.025 \, \text{m}} \]

\[ F_{\text{M}} = 99.8 \, \text{N} \]

Final Answer