Question

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, point a). 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 the class. assume that ( x_1 = 4.5 mathrm{cm}, x_2 = 2.5 mathrm{cm} ), and ( |vec{f}_w| = 53 mathrm{n} ). if the head weighs 53 n, calculate the force exerted by the muscles ( f_m ) using the information in the figure. ( vec{f}_m=) n what is the force ( f_j ) exerted by the pivot on the head?

Explanation:

Step1: Apply torque equilibrium

Torque $\tau = rF\sin\theta$. For rotational equilibrium, $\sum\tau = 0$. Taking torques about point $A$ (the pivot), the torque due to the weight of the head ($\vec{F}_{w}$) is clockwise and the torque due to the muscle force ($\vec{F}_{M}$) is counter - clockwise. Assuming $\theta = 90^{\circ}$ (since the forces are perpendicular to the lever arms), $\tau_{M}=\tau_{w}$. So, $F_{M}x_{1}=F_{w}x_{2}$.

Step2: Solve for $F_{M}$

We know $x_{1} = 4.5\ cm$, $x_{2}=2.5\ cm$, and $F_{w}=53\ N$. From $F_{M}x_{1}=F_{w}x_{2}$, we can express $F_{M}=\frac{F_{w}x_{2}}{x_{1}}$. Substituting the values: $F_{M}=\frac{53\times2.5}{4.5}\approx29.4\ N$.

Step3: Apply force equilibrium

For translational equilibrium, $\sum F_{y}=0$. Let the force exerted by the pivot be $F_{1}$. The forces acting on the head in the vertical direction are $F_{M}$ (upward), $F_{w}$ (downward), and $F_{1}$. So, $F_{1}+F_{M}-F_{w}=0$. Then $F_{1}=F_{w}-F_{M}$.

Step4: Solve for $F_{1}$

Substitute $F_{M}\approx29.4\ N$ and $F_{w} = 53\ N$ into $F_{1}=F_{w}-F_{M}$. So, $F_{1}=53 - 29.4=23.6\ N$.

Answer:

$F_{M}\approx29.4\ N$; $F_{1}\approx23.6\ N$