91Ó°ÊÓ

The angle of inclination of the head from the vertical is:$\mathrm{θ}={40}^{∘}$.

The mass of the head is: m = 4.50 kg.

The distance of center of mass of the head from the pivot point P is: a = 11.0 cm .

The distance of the neck muscles from point P is: b = 1.50 cm .

When a force is applied to a body at a certain distance from the pivot point, the body starts to rotate about the pivot point. The rotational effect of the body is due to the ‘torque’ produced on the body due to the force.

If the distance between the force applied and the pivot point increases, then the torque value increases; otherwise, the torque decreases.

The free-body diagram of the student’s head is given below:

Here, F is the force applied by the neck muscles.

The weight of the student’s head acting downwards is given by,

$$\begin{gathered} W=mg \\ =4.50kg×9.81m/s^2 \\ =44.145N \end{gathered}$$

The formula for the torque produced about the pivot point P is given by,

$$\begin{gathered} \mathrm{τ}=W×a\sin \mathrm{θ} \\ =44.145N×0.11m×\sin {40}^{∘} \\ =3.121N.m \end{gathered}$$

This torque produced is balanced by the neck muscles. Consider the force applied by the neck muscles is F and it acts directly perpendicular to the angle $\mathrm{θ}$of the neck.

So, balancing torques,

$$\begin{gathered} \mathrm{τ}=F×b \\ F=\frac{\mathrm{τ}}{b} \end{gathered}$$

Putting values,

$$\begin{gathered} F=\frac{3.121N.m}{0.015m} \\ =208.07N \end{gathered}$$

Hence, the tension in the neck muscles of the student’s head is 208.07 N.